Question 1

A tower is 50 m high. Its shadow is x metres shorter when the Sun's altitude is 45° than when it is 30°. What is the value of x in metres?

Accepted Answer

Introduction / Context: This tower and shadow problem focuses on how the length of a shadow changes when the Sun's altitude (angle of elevation) changes. You are given the fixed height of the tower and told that the shadow is x metres shorter at an altitude of 45 degrees than at 30 degrees. The goal is to express this difference in shadow lengths in terms of standard trigonometric values involving sqrt(3). Given Data / Assumptions: Height of the tower, H = 50 m. Angle of elevation of the Sun in the first case = 30 degrees. Angle of elevation of the Sun in the second case = 45 degrees. The shadow at 45 degrees is x metres shorter than the shadow at 30 degrees. Level ground and a perfectly vertical tower are assumed. Standard values: tan 30 = 1 / sqrt(3), tan 45 = 1. Concept / Approach: The length of the shadow is given by shadow = height / tan θ. We compute the shadows at 30 degrees and 45 degrees separately and then take the difference. Since the shadow becomes shorter when the Sun is higher (greater angle), the longer shadow will correspond to 30 degrees. The difference (longer minus shorter) is given as x and must be expressed in a factorised form involving sqrt(3). Step-by-Step Solution: Let shadow_30 be the shadow when the angle is 30 degrees. shadow_30 = H / tan 30 = 50 / (1 / sqrt(3)) = 50 * sqrt(3) m. Let shadow_45 be the shadow when the angle is 45 degrees. shadow_45 = H / tan 45 = 50 / 1 = 50 m. Given that shadow_30 - shadow_45 = x. So x = 50√3 - 50 = 50(√3 - 1) m. Verification / Alternative check: We know tan 30 is smaller than tan 45, so shadow_30 must be longer than shadow_45, which we see clearly: approximately 50 * 1.732 = 86.6 m versus 50 m. The difference 86.6 - 50 = 36.6 m, which is consistent with 50(√3 - 1) since (√3 - 1) is about 0.732. This confirms that the algebraic form matches the approximate numeric value. Why Other Options Are Wrong: 50√3 m: This equals the shadow at 30 degrees alone, not the difference. 50(√3 + 1) m: This value is too large and does not represent a simple difference between the two shadows. 50 m: This equals the entire shorter shadow, not the difference of two shadows. 25(√3 - 1) m: This is exactly half of the correct difference and arises if someone accidentally divides by 2. Common Pitfalls: Students may reverse the shadows and incorrectly assume that the shadow at 45 degrees is longer. Another mistake is to use the formula shadow = height * tan θ instead of height / tan θ. Additionally, algebraic factorisation errors, such as not recognising 50√3 - 50 as 50(√3 - 1), can cause confusion when matching with given options. Final Answer: The value of x, the amount by which the shadow is shorter at 45 degrees than at 30 degrees, is 50(√3 - 1) m.

Question 2

The length of the shadow of a tower is √3 times the height of the tower. What is the angle of elevation of the Sun?

Accepted Answer

Introduction / Context: This problem tests your understanding of how the tangent function relates the height of a vertical object and the length of its shadow. The ratio between the shadow length and the height of the tower is given as √3 to 1. From this ratio, you must infer which standard angle of elevation of the Sun gives tan θ equal to 1 / √3. Given Data / Assumptions: Height of the tower = h (a positive constant). Length of the shadow = √3 times the height = √3 h. The tower is vertical and the ground is horizontal. Angle of elevation of the Sun = θ, where tan θ = opposite / adjacent. Opposite side is the tower height and adjacent side is the shadow length. Concept / Approach: Using the right triangle formed by the tower, its shadow and the line from the tip of the shadow to the top, we apply tan θ = height / shadow. The given ratio shadow = √3 * height leads to height / shadow = 1 / √3. We match this with known standard trigonometric values: tan 30 degrees = 1 / √3, tan 45 degrees = 1 and tan 60 degrees = √3. Step-by-Step Solution: Let the height of the tower be h and the length of its shadow be L. Given L = √3 * h. In the right triangle, tan θ = height / shadow = h / L. Substitute L = √3 h to get tan θ = h / (√3 h) = 1 / √3. We know from standard trigonometry that tan 30 degrees = 1 / √3. Therefore θ = 30 degrees. Verification / Alternative check: Recall the basic tan values: tan 30 = 1 / √3, tan 45 = 1, tan 60 = √3. Since we obtained tan θ = 1 / √3, the only compatible acute angle of elevation among the standard choices is 30 degrees. This gives complete agreement with the ratio of shadow length to height provided in the question. Why Other Options Are Wrong: 45°: For 45 degrees, tan 45 = 1, which would require the shadow length and height to be equal, not in the ratio √3 : 1. 60°: For 60 degrees, tan 60 = √3, leading to shadow length = height / √3, which contradicts the given condition. None of these: There is indeed a matching standard angle, 30 degrees, so this is incorrect. 15°: This is not one of the standard angles with tan equal to 1 / √3 and does not satisfy the given ratio. Common Pitfalls: Confusing tan θ with shadow / height instead of height / shadow can lead to inverted ratios. Some students also misremember standard trigonometric values, interchanging tan 30 and tan 60. Always verify with the basic right triangle definitions or remember the common exact values for 30, 45 and 60 degrees. Final Answer: The angle of elevation of the Sun is 30°.

Question 3

Two persons stand on opposite sides of a temple 75 m high and observe the top of the temple at angles of elevation 30° and 60° respectively. What is the distance between the two persons?

Accepted Answer

Introduction / Context: This question involves two observers standing on opposite sides of a tall temple and looking at its top at different angles of elevation. The situation forms two separate right triangles that share the same vertical height (the height of the temple). The aim is to use basic trigonometry to find each observer's horizontal distance from the temple and then add them to get the distance between the two people. Given Data / Assumptions: Height of the temple, h = 75 m. First person sees the top at an angle of elevation θ1 = 30 degrees. Second person sees the top at an angle of elevation θ2 = 60 degrees. The persons are on opposite sides of the temple on level ground. The temple is vertical, forming right angled triangles with the ground. Standard values: tan 30 = 1 / sqrt(3), tan 60 = sqrt(3). Concept / Approach: For each observer, we set up a right triangle with the temple height as the opposite side and the horizontal distance from the temple as the adjacent side. Using tan θ = opposite / adjacent, we can find each horizontal distance. Since the two observers are on opposite sides of the temple in a straight line, the distance between them is the sum of these two distances. Step-by-Step Solution: Let x be the distance of the observer with 60 degree elevation from the temple. Then tan 60 = 75 / x, so sqrt(3) = 75 / x, giving x = 75 / sqrt(3) = 25√3 m. Let y be the distance of the observer with 30 degree elevation from the temple. Then tan 30 = 75 / y, so 1 / sqrt(3) = 75 / y, giving y = 75√3 m. Distance between the two persons = x + y = 25√3 + 75√3 = 100√3 m. Using sqrt(3) ≈ 1.732, 100√3 ≈ 173.2 m. Verification / Alternative check: A quick approximate check: 25√3 is about 43.3 m and 75√3 is about 129.9 m. Their sum is about 173.2 m, which matches the given option exactly. The geometry makes sense: one observer closer to the temple sees a larger angle of elevation (60 degrees), while the more distant observer sees a smaller angle (30 degrees). Why Other Options Are Wrong: 100 m: This ignores the trigonometric ratios and underestimates the total distance. 157.7 m: This is near but does not match the exact value 100√3 ≈ 173.2. 273.2 m: This is too large and would correspond to much greater separation than the actual tan values produce. 150 m: A rounded guess with no proper trigonometric basis in this context. Common Pitfalls: Students may mistakenly subtract the distances instead of adding them, forgetting that the observers are on opposite sides. Another frequent mistake is confusing tan 30 and tan 60, which swaps the distances and leads to incorrect totals. Always draw a simple diagram marking angles and sides, and remember that higher angles of elevation correspond to shorter horizontal distances for a fixed height. Final Answer: The distance between the two persons is 173.2 m (that is, 100√3 m approximately).

Question 4

From the top of a 20 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of the foot of the tower is 45°. What is the height of the tower? (Take √3 = 1.732)

Accepted Answer

Introduction / Context: This question combines angles of elevation and depression to relate the heights of a building and a tower standing on the same horizontal ground. From the top of a known height building, you observe both the top and the foot of a tower at different angles. The objective is to use trigonometry to find the total height of the tower by first obtaining the horizontal distance between the two structures and then using that distance with the angle of elevation. Given Data / Assumptions: Height of the building = 20 m. Angle of depression to the foot of the tower = 45 degrees. Angle of elevation to the top of the tower = 60 degrees. The building and tower stand on the same level ground. Standard values: tan 45 = 1, tan 60 = sqrt(3) (use sqrt(3) ≈ 1.732). Concept / Approach: First, use the angle of depression to find the horizontal distance between the building and the tower. An angle of depression from the horizontal at the top equals the angle of elevation from the foot to the top of the building. With the height of the building and tan 45, we can find the horizontal distance. Then apply the angle of elevation (60 degrees) from the building top to the tower top to get the extra height above the building. Adding this extra height to 20 m gives the total tower height. Step-by-Step Solution: Let the horizontal distance between the building and tower be d. From the 45 degree angle of depression, tan 45 = opposite / adjacent = 20 / d. Since tan 45 = 1, we get 1 = 20 / d, so d = 20 m. Let the height of the tower be H. From the 60 degree angle of elevation: tan 60 = (H - 20) / d. So sqrt(3) = (H - 20) / 20, which implies H - 20 = 20√3. Using √3 ≈ 1.732, H - 20 ≈ 20 × 1.732 = 34.64. Therefore, H ≈ 20 + 34.64 = 54.64 m. Verification / Alternative check: A quick check: With H ≈ 54.64 m, the height difference above the building is about 34.64 m. The ratio (H - 20) / d ≈ 34.64 / 20 ≈ 1.732, which matches tan 60. Also, the distance d = 20 m and building height 20 m give tan 45 = 20 / 20 = 1, which is correct. This confirms that the computed height is consistent with both angles given. Why Other Options Are Wrong: 45.46 m and 45.64 m: These heights are too small and would produce a much smaller angle than 60 degrees from the building top. 54.46 m: Close but slightly off; it does not match the exact use of √3 = 1.732 in the calculation. 50.00 m: This ignores the 60 degree elevation relationships and is not supported by the trigonometric ratios. Common Pitfalls: Students sometimes confuse which segment corresponds to the opposite or adjacent side for each angle. Another common error is to forget that the tower height above the building is H minus 20, not H. Misinterpreting the angle of depression as something different from the corresponding angle of elevation at ground level can also lead to mistakes. A clear labelled diagram is very helpful. Final Answer: The height of the tower is approximately 54.64 m.

Question 5

A ladder leans against a vertical wall making an angle of 60° with the ground. If the foot of the ladder is 4.6 m away from the wall, what is the length of the ladder?

Accepted Answer

Introduction / Context: This is a classic right triangle problem involving a ladder leaning against a vertical wall. The ladder, wall and ground form a right angled triangle where the ladder is the hypotenuse. You are given the angle between the ladder and the ground, as well as the distance from the foot of the ladder to the wall. The question asks for the length of the ladder itself. Given Data / Assumptions: The ladder makes an angle of 60 degrees with the horizontal ground. The foot of the ladder is 4.6 m away from the wall. The wall is vertical and the ground is horizontal, forming a right angle at the foot of the wall. The ladder forms the hypotenuse of the right triangle. Standard value: cos 60 = 1 / 2. Concept / Approach: In a right angled triangle, the cosine of the angle at the base is given by adjacent / hypotenuse, where the adjacent side is along the ground and the hypotenuse is the ladder. Here, we know the adjacent side (4.6 m) and the angle (60 degrees) and need to find the hypotenuse (ladder length). So, hypotenuse = adjacent / cos θ. Using cos 60 = 1 / 2, we can directly compute the length. Step-by-Step Solution: Let L be the length of the ladder. The angle between the ladder and the ground is 60 degrees. By definition, cos 60 = adjacent / hypotenuse = 4.6 / L. Since cos 60 = 1 / 2, we have 1 / 2 = 4.6 / L. Cross multiplying gives L = 4.6 × 2 = 9.2 m. Verification / Alternative check: If L = 9.2 m and the base is 4.6 m, then the height of the ladder on the wall is sqrt(L² - base²) = sqrt(9.2² - 4.6²). This simplifies to sqrt(84.64 - 21.16) = sqrt(63.48) which is about 7.97 m. The sine of 60 degrees is about 0.866, and 0.866 × 9.2 ≈ 7.97 m. This matches the computed height, confirming that the triangle satisfies both sine and cosine relationships for 60 degrees. Why Other Options Are Wrong: 2.3 m: This is only half of the base distance and far too short for a ladder reaching up a wall. 4.6 m: This equals the base distance and would correspond to a 45 degree angle, not 60 degrees. 7.8 m and 8.0 m: These are less than the true hypotenuse computed using cos 60 and do not satisfy the exact trigonometric ratio for 60 degrees. Common Pitfalls: A common mistake is to use sine instead of cosine with the base or to misidentify which side is adjacent and which is opposite. Another error is to compute L = base × cos 60 instead of base / cos 60, which would incorrectly shorten the ladder. Always remember that hypotenuse is longer than either leg and check if your answer is logically larger than the given base distance. Final Answer: The length of the ladder is 9.2 m.

Question 6

The upper part of a tree breaks at a certain height and the broken top touches the ground at a point 10 m from the foot of the tree, making an angle of 60° with the ground. What was the original height of the tree?

Accepted Answer

Introduction / Context: This is a classic broken tree problem that uses right triangle geometry and trigonometry. A tree breaks at some height above the ground, and the upper part falls but remains attached, touching the ground at a known distance from the foot of the tree. The broken part makes a given angle with the ground. Using this information, we must find the original full height of the tree before it broke. Given Data / Assumptions: The upper part of the tree touches the ground at a point 10 m from the foot of the tree. The broken part makes an angle of 60 degrees with the horizontal ground. The tree was originally vertical and the ground is level. The tree broke at a height h above the ground. The broken top segment is straight and has length L. Concept / Approach: When the tree breaks, the portion above the break forms the hypotenuse of a right triangle with the vertical segment from the break point to the ground and the horizontal distance from the foot of the tree to the point where the top touches the ground. The angle between this broken part and the ground is 60 degrees. Using tan 60 = opposite / adjacent, we can find the height at which the tree broke. Then we use Pythagoras to find the length of the broken part and add both to get the original tree height. Step-by-Step Solution: Let h be the height of the break point above the ground. The horizontal distance from the foot of the tree to the point where the top touches the ground is 10 m. The angle between the broken part and the ground at that point is 60 degrees. In the right triangle formed, tan 60 = h / 10. So sqrt(3) = h / 10, giving h = 10√3 m. Now the length of the broken part L is the hypotenuse: L² = h² + 10² = (10√3)² + 10² = 300 + 100 = 400. Thus L = 20 m. Original height of the tree = h + L = 10√3 + 20 = 10(2 + √3) m. Verification / Alternative check: Check numerically: 10√3 is about 17.32 m, so the break point is around that height. The hypotenuse length 20 m is reasonable for a segment leaning at 60 degrees with a base of 10 m, since cos 60 = adjacent / hypotenuse = 10 / 20 = 0.5. This matches the standard value cos 60 = 0.5. Adding the two segments gives approximately 17.32 + 20 = 37.32 m, consistent with the exact expression 10(2 + √3). Why Other Options Are Wrong: 20√3 m and 15√3 m: These represent other combinations of height and hypotenuse but do not match both tan 60 and Pythagoras simultaneously. 10√3 m: This is only the height up to the break point, not the complete original tree height. 10(2 - √3) m: This is much smaller and comes from using an incorrect sign in algebraic manipulation. Common Pitfalls: Students sometimes assume the broken part is equal to the original total height or misplace the 60 degree angle at the foot of the tree instead of at the ground contact of the broken top. Another frequent error is forgetting that the original height is the sum of the vertical part still standing plus the length of the broken segment. Drawing a clear diagram avoids most of these mistakes. Final Answer: The original height of the tree was 10(2 + √3) m.

Question 7

A boat is moving away from an observation tower. When the boat is 50 m from the foot of the tower, the angle of depression of the boat from the top is 60°. After 8 seconds, the angle of depression becomes 30°. Assuming still water, what is the appro…

Accepted Answer

Introduction / Context: This problem combines height and distance with speed, time and distance concepts. A boat moves away from a fixed observation tower. As it moves, the angle of depression from the top of the tower to the boat changes from 60 degrees to 30 degrees over a known time interval. The aim is to use trigonometry to determine the horizontal distances corresponding to these angles and then compute the boat’s speed from the change in distance over time. Given Data / Assumptions: At the first instant, the boat is 50 m from the foot of the tower horizontally. Angle of depression at that instant = 60 degrees. After 8 seconds, the angle of depression becomes 30 degrees. The boat is moving directly away from the tower in still water in a straight line. The tower is vertical and the ground or water surface is horizontal. Angles of depression from the top equal angles of elevation from the boat to the top. Concept / Approach: First, we use the initial angle of 60 degrees and distance 50 m to find the height of the tower using tan 60 = height / base. Then, using the same tower height and the second angle of 30 degrees, we find the new horizontal distance from the tower. Once we have both distances, we calculate how far the boat has travelled in 8 seconds. Finally, we convert this speed from metres per second to kilometres per hour using the conversion factor 1 m/s = 3.6 km/hr. Step-by-Step Solution: Let the height of the tower be h. Initially, base distance x1 = 50 m and angle of elevation is 60 degrees, so tan 60 = h / 50. Thus sqrt(3) = h / 50, giving h = 50√3 m. After 8 seconds, let the new horizontal distance be x2 and angle of elevation be 30 degrees. Then tan 30 = h / x2, so 1 / sqrt(3) = 50√3 / x2. Therefore x2 = 150 m. Distance travelled by the boat in 8 seconds = x2 − x1 = 150 − 50 = 100 m. Speed in m/s = distance / time = 100 / 8 = 12.5 m/s. Convert to km/hr: 12.5 × 3.6 = 45 km/hr. Verification / Alternative check: We can check the tower height with the second distance: using x2 = 150, tan 30 = h / 150 = 1 / sqrt(3), so h = 150 / sqrt(3) = 50√3, consistent with the first calculation. Also, a boat covering 100 m in 8 seconds is traveling at 12.5 m/s, which is a realistic speed for a fast boat, and 12.5 × 3.6 = 45 km/hr aligns exactly with the chosen option. Why Other Options Are Wrong: 33 km/hr and 42 km/hr: These are lower than the correct conversion from 12.5 m/s and do not match the exact calculations. 50 km/hr and 60 km/hr: These are higher than the derived speed and would require the boat to cover more than 100 m in 8 seconds, which contradicts the geometry and time data. Any value not equal to 45 km/hr fails to satisfy both angle conditions and the 8 second interval simultaneously. Common Pitfalls: One common mistake is to misinterpret the initial 50 m as the slant distance instead of the horizontal distance. Another is forgetting that angle of depression equals angle of elevation for the corresponding line of sight. Students may also incorrectly convert from metres per second to kilometres per hour. Always carefully distinguish between vertical height, horizontal distance and the line of sight forming the hypotenuse. Final Answer: The approximate speed of the boat is 45 km/hr.

Question 8

The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 2500 m, what is its speed in km/hr?

Accepted Answer

Introduction / Context: This is a height and distance question involving an aeroplane flying at a fixed altitude. The angle of elevation from a point on the ground to the aeroplane changes as the plane moves, and you are given the time taken for this change. You must find the speed of the aeroplane using trigonometric relationships and speed = distance / time, then convert the result into kilometres per hour. Given Data / Assumptions: Constant height of the aeroplane above ground h = 2500 m. Initial angle of elevation from the observer = 45 degrees. Angle of elevation after 15 seconds = 30 degrees. The aeroplane is flying in a horizontal straight line away from the observer. Standard values: tan 45 = 1, tan 30 = 1 / sqrt(3). Concept / Approach: We model the situation with right angled triangles at two different instants, both having the same vertical height but different horizontal distances from the observer. Using tan θ = height / base, we find the initial and final horizontal distances. Their difference is the horizontal distance travelled during 15 seconds. Dividing by time gives the speed in metres per second, which is then converted to kilometres per hour by multiplying by 3.6. Step-by-Step Solution: Let x1 be the initial horizontal distance when the angle is 45 degrees. tan 45 = 2500 / x1 ⇒ 1 = 2500 / x1 ⇒ x1 = 2500 m. Let x2 be the horizontal distance after 15 seconds when the angle is 30 degrees. tan 30 = 2500 / x2 ⇒ 1 / sqrt(3) = 2500 / x2 ⇒ x2 = 2500√3 m. Horizontal distance travelled in 15 seconds = x2 − x1 = 2500√3 − 2500 = 2500(√3 − 1) m. Speed in m/s = distance / time = 2500(√3 − 1) / 15. Convert to km/hr by multiplying by 3.6: speed = [2500(√3 − 1) / 15] × 3.6 = 600(√3 − 1) km/hr. Verification / Alternative check: Using √3 ≈ 1.732, we get (√3 − 1) ≈ 0.732. Then 600 × 0.732 ≈ 439.2 km/hr, which is a realistic cruising speed for a smaller aircraft. Reversing the process, 439.2 km/hr divided by 3.6 is about 122 m/s. Over 15 seconds, the plane travels roughly 1830 m, which is close to 2500(√3 − 1) ≈ 1830 m, confirming the consistency of the calculation. Why Other Options Are Wrong: 600 km/hr: This assumes a much smaller horizontal distance change and ignores the factor (√3 − 1). 600(√3 + 1) km/hr and 600√3 km/hr: These values are significantly larger and are not supported by the geometry of the two triangles. 500√3 km/hr: This uses an incorrect combination of the height and trig values and does not match the distance travelled in 15 seconds. Common Pitfalls: Common mistakes include swapping the roles of tan 30 and tan 45, or incorrectly assuming the aeroplane is directly overhead at one of the instants. Some learners forget to subtract distances in the correct order to get the distance travelled. Others make unit conversion errors when converting from m/s to km/hr. Always handle each step carefully and keep track of your units. Final Answer: The speed of the aeroplane is 600(√3 - 1) km/hr.

Question 9

The shadow of a tower when the angle of elevation of the Sun is 45° is found to be 10 m longer than when the angle of elevation is 60°. What is the height of the tower?

Accepted Answer

Introduction / Context: This problem again uses the relationship between a tower's height and the length of its shadow for two different angles of elevation of the Sun. We are told that when the angle is 45 degrees, the shadow is 10 m longer than when the angle is 60 degrees. The task is to form an equation involving the unknown tower height and then solve it to obtain a simplified expression that matches one of the given options. Given Data / Assumptions: Height of the tower = H (unknown). Angle of elevation in first case = 45 degrees. Angle of elevation in second case = 60 degrees. Shadow at 45 degrees is 10 m longer than shadow at 60 degrees. The tower is vertical and stands on level ground. Shadow length = height / tan θ for each case. Concept / Approach: Let the shadow when the angle is 45 degrees be S1 and when it is 60 degrees be S2. Using tan 45 = 1 and tan 60 = sqrt(3), we can express S1 and S2 in terms of H. The statement that S1 = S2 + 10 allows us to form an equation and solve for H. The algebraic result can then be simplified to an expression involving sqrt(3) that matches one of the choices. Step-by-Step Solution: Shadow when angle is 45 degrees: S1 = H / tan 45 = H / 1 = H. Shadow when angle is 60 degrees: S2 = H / tan 60 = H / sqrt(3). Given that S1 is 10 m longer than S2: S1 = S2 + 10. So H = H / sqrt(3) + 10. Rearrange: H − H / sqrt(3) = 10. Factor H: H(1 − 1 / sqrt(3)) = 10. Compute the bracket: 1 − 1 / sqrt(3) = (sqrt(3) − 1) / sqrt(3). Thus H × (sqrt(3) − 1) / sqrt(3) = 10 ⇒ H = 10 sqrt(3) / (sqrt(3) − 1). Multiply numerator and denominator by (sqrt(3) + 1): H = 10 sqrt(3)(sqrt(3) + 1) / (3 − 1) = 5 sqrt(3)(sqrt(3) + 1). So H = 5(3 + sqrt(3)) = 15 + 5√3 m, written as 5√3 + 15 m. Verification / Alternative check: Approximate √3 ≈ 1.732. Then H ≈ 5(3 + 1.732) = 5 × 4.732 = 23.66 m. Now S1 = H ≈ 23.66 m and S2 = H / √3 ≈ 23.66 / 1.732 ≈ 13.66 m. The difference S1 − S2 ≈ 23.66 − 13.66 = 10 m, which matches the given condition exactly. Why Other Options Are Wrong: 5(√3 − 1) m and 10(√3 − 1) m: These are much smaller heights and do not produce a 10 m difference between shadows. 5(√3 + 1) m: This is again too small; it would give a shorter shadow and a smaller difference. 10(√3 + 1) m: This is larger but still does not satisfy the exact equation H − H / √3 = 10. Common Pitfalls: Students often write the difference equation in the wrong order, using S2 − S1 = 10 instead of S1 − S2 = 10. Another typical error is to forget to factor out H correctly or to mishandle the rationalisation step involving √3. Carefully keeping track of algebraic manipulations and double checking with an approximate numerical check helps avoid such mistakes. Final Answer: The height of the tower is 5√3 + 15 m.

Question 10

Two men stand on opposite sides of a tower and measure the angles of elevation of the top of the tower as 30° and 45° respectively. If the height of the tower is 50 m (take √3 = 1.73), what is the distance between the two men?

Accepted Answer

Introduction / Context: In this problem, two observers stand on opposite sides of a tower and view its top under different angles of elevation. You are given the height of the tower and must find the distance between the two observers. This is a direct application of the tangent function for each right triangle formed with the tower height as the common vertical side. Given Data / Assumptions: Height of the tower, h = 50 m. Angle of elevation from the first man = 30 degrees. Angle of elevation from the second man = 45 degrees. The two men are on opposite sides of the tower on the same straight line. The tower is vertical and the ground is horizontal. Approximate value: √3 ≈ 1.73. Concept / Approach: We form two right triangles with the tower height as the common opposite side and different adjacent sides representing the distances from each man to the tower. Using tan θ = opposite / adjacent, we find each distance. Because the men are on opposite sides, the distance between them is the sum of these two distances. Finally, we substitute √3 ≈ 1.73 to get a decimal value that matches one of the options. Step-by-Step Solution: Let d1 be the distance of the man who sees the top at 30 degrees. tan 30 = 50 / d1 ⇒ 1 / √3 = 50 / d1 ⇒ d1 = 50√3 m. Let d2 be the distance of the man who sees the top at 45 degrees. tan 45 = 50 / d2 ⇒ 1 = 50 / d2 ⇒ d2 = 50 m. Distance between the two men = d1 + d2 = 50√3 + 50. Using √3 ≈ 1.73, 50√3 ≈ 50 × 1.73 = 86.5 m. Thus distance ≈ 86.5 + 50 = 136.5 m. Verification / Alternative check: Check each triangle separately. For the 30 degree observer: tan 30 ≈ 0.577, and 50 / 0.577 ≈ 86.6 which matches 50√3 ≈ 86.5. For the 45 degree observer: tan 45 = 1, so base = height = 50 m, which is consistent. Adding 86.5 and 50 gives 136.5 m, confirming our result is accurate and consistent with all given data. Why Other Options Are Wrong: 50√3 m: This is only the distance of one man from the tower, not the distance between the two men. 100√3 m: This would correspond to each man being 50√3 m away on the same side, which contradicts the 45 degree angle for one observer. 135.5 m: This is a near miss but does not match the exact computation 50√3 + 50 with √3 = 1.73. 120 m: A rough guess with no basis in the correct trigonometric formulae. Common Pitfalls: Learners often subtract the distances instead of adding them because they misinterpret the phrase "opposite sides." It is also easy to confuse tan 30 and tan 60 or to use an inaccurate approximation for √3 without checking. Drawing a diagram and labelling each distance clearly helps avoid such confusion. Final Answer: The distance between the two men is 136.5 m.

Question 11

A 25 m long ladder rests against a vertical wall. Initially, the foot of the ladder is 7 m away from the wall. If the top of the ladder slides down 4 m, how far will the foot of the ladder move away from the wall?

Accepted Answer

Introduction / Context: This problem involves a ladder sliding along a vertical wall while its foot moves along the ground. The ladder length remains constant, so the situation is governed by the Pythagoras theorem. You are given the initial position and then told that the top has slid down by 4 m. The goal is to determine the new position of the foot and how far it has moved from its original place. Given Data / Assumptions: Length of the ladder L = 25 m (constant). Initially, the foot of the ladder is 7 m away from the wall. Initially, the height reached on the wall is unknown but can be found. The top of the ladder slides down by 4 m. The wall is vertical, the ground is horizontal, and the ladder is straight. Concept / Approach: The ladder, wall and ground form a right angled triangle where the ladder is the hypotenuse. Initially, we use Pythagoras theorem to find the height of the ladder on the wall. After the top slides down by 4 m, we again apply Pythagoras with the new height and the fixed ladder length to compute the new distance from the wall. The movement of the foot is the difference between the new base distance and the original 7 m. Step-by-Step Solution: Initial situation: base distance x1 = 7 m, ladder length L = 25 m. Let initial height be h1. By Pythagoras: h1² + 7² = 25². So h1² + 49 = 625 ⇒ h1² = 576 ⇒ h1 = 24 m. The top slides down 4 m, so new height h2 = 24 − 4 = 20 m. Let the new base distance be x2. Again, h2² + x2² = L². So 20² + x2² = 25² ⇒ 400 + x2² = 625 ⇒ x2² = 225. Thus x2 = 15 m. Movement of the foot = x2 − x1 = 15 − 7 = 8 m. Verification / Alternative check: Check both positions with the fixed ladder length: Initially, sqrt(7² + 24²) = sqrt(49 + 576) = sqrt(625) = 25 m. After movement, sqrt(15² + 20²) = sqrt(225 + 400) = sqrt(625) = 25 m again. This confirms that the ladder length remains consistent in both positions. The difference in base distances is exactly 8 m, which aligns with our algebraic result. Why Other Options Are Wrong: 5 m, 6 m, 9 m, 10 m: These values do not satisfy the right triangle constraints when combined with the 4 m vertical movement and a fixed 25 m ladder. None gives both initial and final triangles with hypotenuse 25 m and correct heights. Only 8 m exactly maintains the constant ladder length and the stated vertical slide distance. Common Pitfalls: A frequent error is to treat the ladder movement as linear and assume the foot moves the same distance as the top. That is incorrect because the ladder is rotating. Another mistake is failing to recompute the new base distance using Pythagoras theorem after the top slides down. Always remember that the relationship between height and base for a fixed hypotenuse is non linear. Final Answer: The foot of the ladder will move away from the wall by 8 m.

Question 12

From a point 30 m away from the foot of a flag post, the angles of elevation of the bottom and the top of a flag on the post are 30° and 45° respectively. Taking √3 = 1.732, what is the height of the flag alone?

Accepted Answer

Introduction / Context: This problem features a flag mounted on a flag post. From a point on the ground 30 m away, the angles of elevation to the bottom and top of the flag are given. By modelling two right triangles that share the same base distance, we can find the heights of the bottom and top of the flag above the ground and then subtract to get the height of the flag itself. Given Data / Assumptions: Horizontal distance from the observer to the foot of the flag post = 30 m. Angle of elevation to the bottom of the flag = 30 degrees. Angle of elevation to the top of the flag = 45 degrees. The post is vertical and the ground is horizontal. Take √3 ≈ 1.732. Concept / Approach: The line of sight to the bottom of the flag and the line of sight to the top form two right angled triangles with the same base (30 m). Let the height of the bottom of the flag above ground be h1 and the height of the top be h2. Using tan 30 and tan 45, we can find h1 and h2. The height of the flag alone is h2 − h1. We then approximate using √3 ≈ 1.732 to obtain a decimal answer. Step-by-Step Solution: Let h1 be the height of the bottom of the flag. tan 30 = h1 / 30 ⇒ 1 / √3 = h1 / 30 ⇒ h1 = 30 / √3 = 10√3 m. Let h2 be the height of the top of the flag. tan 45 = h2 / 30 ⇒ 1 = h2 / 30 ⇒ h2 = 30 m. Height of the flag alone = h2 − h1 = 30 − 10√3 m. Using √3 ≈ 1.732, 10√3 ≈ 17.32 m. Therefore, height of the flag ≈ 30 − 17.32 = 12.68 m. Verification / Alternative check: Check each height separately: h1 ≈ 17.32 m and h2 = 30 m. For the bottom, tan 30 ≈ 0.577 and h1 / 30 ≈ 17.32 / 30 ≈ 0.577, matching tan 30. For the top, tan 45 = 1 and h2 / 30 = 30 / 30 = 1, matching tan 45. The difference 12.68 m therefore reliably represents the vertical height occupied by the flag itself. Why Other Options Are Wrong: 12√3 m: This is about 20.78 m, which is too large given that the whole post plus flag height seen at 45 degrees is only 30 m. 15 m: This is a rough guess and does not match 30 − 10√3 exactly. 14.32 m: This would require different numeric relationships and does not stem from the correct tan values. 10√3 m: This corresponds to the height of the bottom of the flag, not to the height of the flag itself. Common Pitfalls: A common error is to treat 30 m as the height or to forget to subtract the two heights, reporting 30 m or 10√3 m directly. Another pitfall is reversing which angle corresponds to which height or mixing up tan 30 and tan 45. Careful reading of the question and a simple diagram usually prevent such confusion. Final Answer: The height of the flag alone is approximately 12.68 m.

Question 13

A pilot flying at an altitude of 200 m observes two points on opposite banks of a river. The angles of depression of the two points are 45° and 60° respectively. What is the width of the river?

Accepted Answer

Introduction / Context: This question describes an aeroplane flying horizontally over a river while a pilot observes two points on opposite banks at different angles of depression. Because both observations are made from the same altitude, we get two right triangles sharing the same vertical height but with different horizontal distances from the point on the ground directly below the aeroplane. The width of the river is the distance between these two ground points. Given Data / Assumptions: Altitude of the aeroplane above the river surface = 200 m. Angles of depression to the two opposite banks = 45 degrees and 60 degrees. The river banks lie on a straight line on either side of the point vertically below the aeroplane. Angles of depression equal the corresponding angles of elevation from the river banks. Standard values: tan 45 = 1, tan 60 = sqrt(3). Concept / Approach: Let the point vertically below the aeroplane on the water surface be P. Let A and B be the points on the nearer and farther banks respectively, such that angles of depression are 60 degrees to A and 45 degrees to B. Then AP and BP are horizontal distances. For each right triangle, tan θ = height / base. We find AP and BP and then sum them to get the total width AB = AP + BP, because the banks lie on opposite sides of P. Step-by-Step Solution: Let AP be the horizontal distance to the nearer bank where the angle of depression is 60 degrees. Then tan 60 = 200 / AP ⇒ sqrt(3) = 200 / AP ⇒ AP = 200 / sqrt(3) m. Let BP be the horizontal distance to the farther bank where the angle of depression is 45 degrees. Then tan 45 = 200 / BP ⇒ 1 = 200 / BP ⇒ BP = 200 m. Width of the river AB = AP + BP = 200 / √3 + 200. Thus, width = (200 + 200/√3) m. Verification / Alternative check: Using √3 ≈ 1.732, AP ≈ 200 / 1.732 ≈ 115.47 m and BP = 200 m. So the width is about 315.47 m. Both triangles use the same vertical height 200 m. For AP, tan 60 ≈ 200 / 115.47 ≈ 1.732, which matches sqrt(3). For BP, tan 45 = 200 / 200 = 1, as expected. The sum of the distances matches the derived algebraic expression. Why Other Options Are Wrong: (200 − 200/√3) m: This would be the difference of distances, not the sum, and does not represent the full river width when the plane is above a point between the banks. 400√3 m and (400/√3) m: These values are much larger or smaller than what the tan relations provide and would not maintain the correct angle relationships. 200√3 m: This corresponds to a single base distance with tan 60, not the combined width of both banks. Common Pitfalls: Students may mistakenly subtract distances instead of adding them by misinterpreting the location of the aeroplane relative to the river banks. Another error is to confuse which angle corresponds to the nearer bank and which to the farther bank. Remember that the larger angle of depression corresponds to the closer point (shorter base distance). Drawing a simple diagram is very helpful in avoiding such mistakes. Final Answer: The width of the river is (200 + 200/√3) m.

Question 14

A telegraph post is bent at a point above the ground. Its top just touches the ground at a distance of 8√3 m from its foot and makes an angle of 30° with the horizontal. What is the height of the post in metres?

Accepted Answer

Introduction / Context: This question is similar to the classic broken tree problem. A telegraph post is bent somewhere above the ground and its top touches the ground at some horizontal distance from the foot of the post. The broken part makes a known angle with the horizontal. We must determine the total original height of the post by applying basic trigonometry and Pythagoras theorem. Given Data / Assumptions: The top of the post touches the ground at a distance of 8√3 m from its foot. The broken top makes an angle of 30 degrees with the horizontal. The post was originally vertical and straight. It is bent at a point h metres above the ground. The length of the broken part is L. Concept / Approach: When the post bends, the section above the bend becomes the hypotenuse of a right triangle whose base is the horizontal distance from the foot to the contact point and whose height is the vertical distance from the bend to the ground. Since the angle between the broken part and the ground is 30 degrees, we can use tan 30 = opposite / adjacent to find the height of the bend. Then, using Pythagoras theorem with the base 8√3 and this height, we compute the length of the broken part and add the two vertical segments to get the total height of the post. Step-by-Step Solution: Let h be the height of the bend above the ground. The horizontal distance from the foot to where the top touches the ground is 8√3 m. The angle between the broken part and the ground is 30 degrees. Using tan 30 = h / (8√3), we have 1 / √3 = h / (8√3). So h = (8√3) × (1 / √3) = 8 m. Let L be the length of the broken part (hypotenuse of the triangle). L² = h² + (8√3)² = 8² + 64 × 3 = 64 + 192 = 256. Hence L = 16 m. Total height of the telegraph post = h + L = 8 + 16 = 24 m. Verification / Alternative check: Quickly check with sine or cosine: For an angle of 30 degrees, cos 30 = adjacent / hypotenuse = (8√3) / 16 = (8√3) / 16 = √3 / 2, which matches cos 30. Also, sin 30 = opposite / hypotenuse = 8 / 16 = 1 / 2, which matches sin 30. Both checks confirm that our calculated lengths are consistent with the given angle. Why Other Options Are Wrong: 12 m, 16 m, 18 m, 20 m: These values are all less than the sum of the two computed segments h and L and cannot satisfy both the angle condition and Pythagoras theorem simultaneously. Only 24 m correctly matches the geometry and trigonometry of the problem. Common Pitfalls: Sometimes students confuse the 30 degree angle as being at the foot of the post instead of at the point where the top touches the ground. Another mistake is to forget that the total height includes both the vertical segment up to the bend and the length of the broken segment. Drawing the right triangle and marking the angle helps clarify which side is which. Final Answer: The height of the telegraph post is 24 m.

Question 15

The angles of elevation of the top of a tower from two points on the same side of the tower and in a straight line with it, at distances 4 m and 9 m from its base, are complementary. What is the height of the tower?

Accepted Answer

Introduction / Context: This problem uses complementary angles of elevation to the top of a tower from two points at known distances from its base. Two right triangles share the same vertical height but have different bases. You must use the property that tan θ and tan(90° − θ) are reciprocal to each other to form an equation and then solve for the height of the tower. Given Data / Assumptions: Distances of the two observation points from the tower base are 4 m and 9 m. The angles of elevation from these two points are complementary (their sum is 90 degrees). The tower is vertical and stands on horizontal ground. Let the two angles be α and 90° − α respectively. In a right triangle, tan(90° − α) = 1 / tan α. Concept / Approach: Let the height of the tower be H. Then tan α = H / 4 for the nearer point and tan(90° − α) = H / 9 for the farther point. Because tan(90° − α) = 1 / tan α, we can construct a simple equation involving H and the two base distances. Solving this equation yields the height of the tower, which can be checked against the given options. Step-by-Step Solution: Let H be the height of the tower. From the point at distance 4 m, tan α = H / 4. From the point at distance 9 m, angle of elevation is 90° − α, so tan(90° − α) = H / 9. But tan(90° − α) = 1 / tan α. Thus H / 9 = 1 / (H / 4) = 4 / H. Cross multiplying: H² = 36. So H = 6 m (taking the positive root for height). Verification / Alternative check: If H = 6 m, then tan α = 6 / 4 = 3 / 2 = 1.5. So tan(90° − α) must be 1 / 1.5 = 2 / 3. From the second point, tan(90° − α) = H / 9 = 6 / 9 = 2 / 3. The two expressions match, confirming that the angles are indeed complementary and that H = 6 m is correct. Why Other Options Are Wrong: 4 m, 5 m, 7 m, 9 m: Substituting any of these values into the relationships tan α = H / 4 and tan(90° − α) = H / 9 destroys the reciprocal property required by complementary angles. Only 6 m satisfies H / 9 = 4 / H while keeping H positive and giving consistent triangles. Common Pitfalls: A common mistake is to assume that complementary angles mean the sum of their tangents is 1, which is incorrect. The correct property is that tan(90° − α) is the reciprocal of tan α. Another error is to confuse the distances 4 m and 9 m or to place them incorrectly in the triangles. A neat diagram and careful application of the tan definitions help avoid these issues. Final Answer: The height of the tower is 6 m.

Question 16

A helicopter flying at an altitude of 1500 m observes two ships sailing towards it in the same straight line. The angles of depression of the ships are 60° and 30° respectively. What is the distance between the two ships in metres?

Accepted Answer

Introduction / Context: Here a helicopter hovers at a fixed altitude and observes two ships that are on the same straight line, sailing towards the point directly beneath the helicopter. The angles of depression to the two ships are 60 degrees and 30 degrees. Using these angles and the fixed altitude, we can find each ship's horizontal distance from the helicopter's projection on the water surface and then compute how far apart the ships are from each other. Given Data / Assumptions: Altitude of the helicopter above sea level = 1500 m. Angles of depression to the two ships = 60 degrees and 30 degrees. The ships are on the same straight line and on the same side relative to the helicopter's vertical projection. Angles of depression equal corresponding angles of elevation from the ships. Standard values: tan 60 = sqrt(3), tan 30 = 1 / sqrt(3). Concept / Approach: Let the point directly below the helicopter on the water surface be P. Let the nearer ship be at distance x1 from P with angle of elevation 60 degrees and the farther ship at distance x2 from P with angle of elevation 30 degrees. For each triangle, tan θ = height / base. We compute x1 and x2 from the known height 1500 m. The distance between the ships is the absolute difference |x2 − x1| because they lie along the same horizontal line on the same side. Step-by-Step Solution: Let x1 be the horizontal distance to the ship seen at angle of depression 60 degrees. tan 60 = 1500 / x1 ⇒ sqrt(3) = 1500 / x1 ⇒ x1 = 1500 / sqrt(3) m. Let x2 be the horizontal distance to the ship seen at angle of depression 30 degrees. tan 30 = 1500 / x2 ⇒ 1 / sqrt(3) = 1500 / x2 ⇒ x2 = 1500√3 m. Distance between the two ships = x2 − x1 (since x2 > x1). So distance = 1500√3 − 1500 / √3. Factor: 1500(√3 − 1 / √3) = 1500((3 − 1) / √3) = 1500 × 2 / √3 = 3000 / √3. Note that 3000 / √3 = 1000 × (3 / √3) = 1000√3 m. Verification / Alternative check: Using √3 ≈ 1.732, x1 ≈ 1500 / 1.732 ≈ 866 m, and x2 ≈ 1500 × 1.732 ≈ 2598 m. Their difference is about 1732 m, which is close to 1000√3 ≈ 1732 m. Both triangles are consistent with tan 60 and tan 30 using the same height of 1500 m, which confirms our calculations. Why Other Options Are Wrong: 1000/√3 m and 500/√3 m: These values are much smaller than the actual separation determined by the geometry. 500√3 m: This is roughly half the correct separation and does not correspond to the difference between 1500√3 and 1500 / √3. 1500√3 m: This is just the distance from P to the farther ship, not the distance between the two ships. Common Pitfalls: Mistakes often arise from confusing which ship corresponds to which angle. The nearer ship must have the larger angle (60 degrees) and thus the smaller base distance. Another trap is to add the two distances instead of subtracting them, which would imply the ships are on opposite sides of the helicopter, contradicting the statement that they sail towards it along the same direction. Careful interpretation of the problem text is essential. Final Answer: The distance between the two ships is 1000√3 m.

Question 17

From a point 30 m above the surface of a lake, the angle of elevation of an aeroplane is 30° and the angle of depression of its image in the water is 60°. What is the height of the aeroplane above the water surface?

Accepted Answer

Introduction / Context: This problem uses the idea of reflection in water together with angles of elevation and depression. An observer stands at a point above the surface of a lake and observes both an aeroplane in the sky and its image reflected in the water. Using the angles to the aeroplane and its image, we need to determine the actual height of the aeroplane above the water surface. Given Data / Assumptions: The observer is 30 m above the water surface. The angle of elevation from the observer to the aeroplane is 30 degrees. The angle of depression from the observer to the image of the aeroplane in the water is 60 degrees. The image of the aeroplane in the water is as far below the water level as the plane is above it. The lake surface is horizontal and acts as a mirror. Concept / Approach: Let the aeroplane be at height H above the water surface and at horizontal distance x from the vertical line through the observer. Its image will be at height −H (below the surface) but at the same horizontal distance x. The vertical distance from observer (30 m above water) to the plane is H − 30, and to the image is 30 + H. Using tan 30 = (H − 30) / x and tan 60 = (30 + H) / x, we can solve the system of equations for H and x, then report H as the height above water. Step-by-Step Solution: Let the height of the aeroplane above the water surface be H. Let the horizontal distance from the observer's vertical projection to the aeroplane be x. For the aeroplane: tan 30 = (H − 30) / x ⇒ 1 / √3 = (H − 30) / x. So H − 30 = x / √3. (1) For the image: angle of depression is 60 degrees, so tan 60 = (30 + H) / x ⇒ √3 = (30 + H) / x. So 30 + H = x√3. (2) From (1), x = √3(H − 30). Substitute into (2): 30 + H = √3(H − 30) × √3 = 3(H − 30). Thus 30 + H = 3H − 90 ⇒ 30 + 90 = 3H − H ⇒ 120 = 2H ⇒ H = 60 m. Verification / Alternative check: If H = 60 m, then from (1), H − 30 = 30 m. So x / √3 = 30 ⇒ x = 30√3. From (2), 30 + H = 90, and x√3 = 30√3 × √3 = 30 × 3 = 90, which matches perfectly. Tan 30 = 30 / (30√3) = 1 / √3 and tan 60 = 90 / (30√3) = 3 / √3 = √3, confirming that both angles are correct. Why Other Options Are Wrong: 45 m and 50 m: These heights do not satisfy both tan 30 and tan 60 relations simultaneously when substituted into the system. 75 m and 90 m: These heights give inconsistent distances or angles relative to the given 30 m observer height and the stated angles. Only 60 m yields a consistent pair of right triangles with correct angles and distances. Common Pitfalls: Many students forget that the distance from the observer to the image is 30 + H, not H − 30, because the image is below the water surface. Another mistake is to mismatch the angles with the wrong vertical distances. Drawing a vertical axis showing the observer, water surface, aeroplane and its image often helps keep track of the correct distances and signs. Final Answer: The height of the aeroplane above the water surface is 60 m.

Question 18

The angles of elevation of the top of a tree 220 m high from two points in the same vertical plane are 30° and 45°. What is the distance between these two points in metres?

Accepted Answer

Introduction / Context: In this problem, two observers are in the same vertical plane with a tall tree. They see the top of the tree under different angles of elevation, 30 degrees and 45 degrees. Given the height of the tree, we must use trigonometry to find how far each observer is from the tree and then determine the distance between the two observers. Given Data / Assumptions: Height of the tree, h = 220 m. Angles of elevation from the two points = 30 degrees and 45 degrees. The two points lie in the same vertical plane with the tree and on the same side of it. The ground is horizontal and the tree is vertical. Standard values: tan 30 = 1 / sqrt(3), tan 45 = 1. Concept / Approach: Let the distances from the tree base to the two points be d1 and d2, with d1 > d2, corresponding to angles 30 degrees and 45 degrees respectively (smaller angle means farther distance). We use tan θ = height / base to compute d1 and d2. Then the distance between the two points is simply d1 − d2. Finally, using √3 ≈ 1.732, we approximate the numeric value to match one of the options. Step-by-Step Solution: Let d1 be the distance to the point where the angle of elevation is 30 degrees. tan 30 = 220 / d1 ⇒ 1 / √3 = 220 / d1 ⇒ d1 = 220√3 m. Let d2 be the distance to the point where the angle of elevation is 45 degrees. tan 45 = 220 / d2 ⇒ 1 = 220 / d2 ⇒ d2 = 220 m. Distance between the two points = d1 − d2 = 220√3 − 220 = 220(√3 − 1) m. Using √3 ≈ 1.732, √3 − 1 ≈ 0.732. Thus distance ≈ 220 × 0.732 ≈ 161.04 m (rounded to 161.05 m). Verification / Alternative check: Check d1: 220√3 ≈ 220 × 1.732 ≈ 381 m. Then tan 30 ≈ 220 / 381 ≈ 0.577, matching 1 / √3. Check d2: 220 / 220 = 1, matching tan 45. The difference 381 − 220 ≈ 161 m aligns with the approximate value obtained by the algebraic expression 220(√3 − 1). Everything is consistent with the given angles and height. Why Other Options Are Wrong: 193.22 m and 176.12 m: These overestimate the separation and are not supported by the exact expression 220(√3 − 1). 144.04 m and 150.00 m: These underestimate the distance and again do not correspond to 220(√3 − 1). Only 161.05 m matches the accurate computation using standard trigonometric values. Common Pitfalls: One common error is to add d1 and d2 instead of subtracting them, which would suggest that the points are on opposite sides of the tree. Another mistake is to confuse tan 30 with tan 60 or to misassign the larger distance to the larger angle. Remember that for the same height, a smaller angle of elevation always corresponds to a larger horizontal distance. Final Answer: The distance between the two points is approximately 161.05 m.

Question 19

From the top and from the bottom of a building, the angles of elevation of the top of a vertical tower 72 metres high are 30° and 60° respectively. What is the height, in metres, of the building?

Accepted Answer

Introduction / Context: This question is based on heights and distances, where angles of elevation are used to relate vertical heights and horizontal distances in right angled triangles. We are given a tower of known height and a building of unknown height, and two different angles of elevation of the tower top observed from the bottom and from the top of the building. Using simple trigonometry with tangent ratios, we can form equations and determine the height of the building. Given Data / Assumptions: The height of the vertical tower is 72 metres. From the bottom of the building, the angle of elevation of the top of the tower is 60 degrees. From the top of the building, the angle of elevation of the top of the tower is 30 degrees. The building and the tower stand on the same horizontal ground. The horizontal distance between the building and the tower is the same for both observations. Concept / Approach: In height and distance problems, the tangent of an angle of elevation is equal to vertical height divided by horizontal distance. Here we use tan 60 degrees and tan 30 degrees. Let the height of the building be h metres and the horizontal distance between the building and the tower be x metres. From the two angles, we obtain two equations in terms of h and x and then solve them simultaneously to find h. Step-by-Step Solution: Step 1: From the bottom of the building, tan 60° = 72 / x, so √3 = 72 / x and therefore x = 72 / √3 = 24√3 metres. Step 2: From the top of the building, the vertical difference between the top of the tower and the top of the building is 72 − h. Step 3: The angle of elevation from the top is 30°, so tan 30° = (72 − h) / x. Therefore 1 / √3 = (72 − h) / x. Step 4: Substitute x = 24√3 into the equation: 1 / √3 = (72 − h) / (24√3). Step 5: Multiply both sides by 24√3 to get 24 = 72 − h, so h = 72 − 24 = 48 metres. Verification / Alternative check: Check with both angles. With x = 24√3 and h = 48, the vertical difference from bottom is 72, so tan 60° = 72 / (24√3) = 72 / (24 * 1.732) which simplifies to √3, correct. The vertical difference from top is 72 − 48 = 24, so tan 30° = 24 / (24√3) = 1 / √3, correct. Therefore the value h = 48 metres is consistent for both observations. Why Other Options Are Wrong: Option 42 would give a vertical difference of 30 metres from the top, which does not produce tan 30° correctly. Option 20√3 and option 24√3 are heights that do not satisfy both angle equations simultaneously. Option 36 also fails when checked with tan 30° and tan 60° conditions. Only 48 metres satisfies both trigonometric relationships exactly. Common Pitfalls: A common mistake is to interchange the angles, or to forget that the vertical distance from the top of the building to the top of the tower is 72 minus h and not 72 plus h. Another frequent error is mishandling √3 when simplifying expressions such as 72 / √3. Students sometimes also confuse angle of elevation with angle of depression, although here only angles of elevation are used. Final Answer: The height of the building is 48 metres.

Question 20

From a point on the ground, the angle of elevation of an aeroplane is 60 degrees. After flying for 30 seconds in a straight horizontal line at a constant height of 4500 metres, the angle of elevation becomes 30 degrees. What is the speed of the aero…

Accepted Answer

Introduction / Context: This problem involves an aeroplane flying in a straight horizontal path at a constant height above the ground. Two different angles of elevation of the aeroplane are observed from the same point on the ground at two different times. Using trigonometry and the fact that the height is constant, we can calculate the horizontal distances at both instants and thus determine the speed of the aeroplane in metres per second. Given Data / Assumptions: Initial angle of elevation of the aeroplane from a fixed ground point is 60 degrees. After 30 seconds, the angle of elevation from the same point becomes 30 degrees. The aeroplane flies at a constant height of 4500 metres above the ground. The aeroplane moves in a straight horizontal line. We neglect curvature of the earth or wind effects, and treat the path as a right angled triangle situation at each instant. Concept / Approach: In height and distance questions, for a right angled triangle, tan of the angle of elevation is equal to the vertical height divided by the horizontal distance from the observer. Let the horizontal distances at the two instants be x1 and x2. Using tan 60 degrees and tan 30 degrees, we express x1 and x2 in terms of the fixed height. The aeroplane covers a horizontal distance of (x2 − x1) in 30 seconds, so the speed is (x2 − x1) divided by 30, expressed in metres per second. Step-by-Step Solution: Step 1: At the first instant, tan 60° = 4500 / x1, so √3 = 4500 / x1, giving x1 = 4500 / √3 = 1500√3 metres. Step 2: After 30 seconds, tan 30° = 4500 / x2, so 1 / √3 = 4500 / x2, giving x2 = 4500√3 metres. Step 3: Horizontal distance covered in 30 seconds is x2 − x1 = 4500√3 − 1500√3 = 3000√3 metres. Step 4: Speed of the aeroplane = distance / time = (3000√3) / 30 = 100√3 metres per second. Verification / Alternative check: An approximate numerical check helps. Take √3 approximately equal to 1.732. Then 100√3 is about 173.2 metres per second. In 30 seconds the aeroplane would travel about 5196 metres horizontally. The initial distance is around 2598 metres and the final around 7794 metres, whose difference is very close to 5196 metres, matching our calculation. The angles of 60 degrees and 30 degrees are also consistent with the computed distances and constant height of 4500 metres. Why Other Options Are Wrong: Option 50√3 corresponds to half the correct horizontal speed and would imply a much smaller change in horizontal distance, inconsistent with the change in angle. Options 200√3 and 300√3 imply much larger speeds, which would overshoot the required change in distance for the given height. Option 150√3 is also larger than the derived value and does not fit the trigonometric relations for both angles. Common Pitfalls: Many learners incorrectly assume that the aeroplane is approaching the observer when the angle decreases, but here the angle of elevation reduces from 60 degrees to 30 degrees so the plane is moving away. Another pitfall is forgetting to convert time or speed units, but in this problem everything is already in seconds and metres. Errors in handling √3 or in subtracting x1 from x2 also lead to wrong answers. Final Answer: The speed of the aeroplane is 100√3 metres per second.

Height and Distance Questions