Question 1

Jack takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

Accepted Answer

Average speed = total distance / total time Total distance covered = 6 miles; total time = 45 minutes = 0.75 hours Average speed = 6/ 0.75 = 8 miles/hour

Question 2

The top and bottom of a tower were seen to be at angles of depression 30° and 60° from the top of a hill of height 100 m. Find the height of the tower ?

Accepted Answer

66.6 mts

Question 3

A flagstaff 17.5 m high casts a shadow of length 40.25 m. What will be the height of a building, which casts a shadow of length 28.75 m under similar conditions ?

Accepted Answer

Let the required height of the building be x meter More shadow length, More height(direct proportion) Hence we can write as (shadow length) 40.25 : 28.75 :: 17.5 : x ⇒ 40.25 × x = 28.75 × 17.5 ⇒ x = 28.75 × 17.5/40.25 = 2875 × 175/40250 = 2875 × 7/1610 = 2875/230 = 575/46 = 12.5 cm

Question 4

A ladder is resting against a vertical wall and its bottom is 2.5 m away from the wall. If it slips 0.8 m down the wall, then its bottom will move away from the wall by 1.4 m. What is the length of the ladder?

Accepted Answer

6.5 m

Question 5

The height and the slant height of a right circular cone are 24 cm and 25 cm, respectively. Considering π as 22/7 , find the curved surface area of the said cone.

Accepted Answer

550 cm2

Question 6

Two men standing on same side of a pillar 75 metre high, observe the angles of elevation of the top of the pillar to be 30° and 60° respectively the distance between two men is

Accepted Answer

100 3m

Question 7

There are two parallel streets each directed north to south. A person in the first street travelling from south to north wishes to take the second street which is on his right side. At some place, he makes a 150 deg turn to the right and he travels …

Accepted Answer

Initially the person is travelling from south to north i.e. D to A He takes 150 deg right turn and moves AB distance and then he takes 60 deg left turn travels BC AB = 20km/hr * 15/60 hr = 5km BC = 30 * 20/60 = 10 km We know that distance between both the streets is DC = DB + BC DB = AB cos 60o= 5. ½ =2.5 km So the distance between streets = 12.5 km

Question 8

The top of a broken tree touches the ground at a distance of 15 m from its base. If the tree is broken at a height of 8 m from the ground, then the actual height of the tree is

Accepted Answer

25 m

Question 9

A ladder 13 m long reaches a window which is 12 m above the ground on side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to each a window 5m high, then the width of the street is :

Accepted Answer

17 m

Question 10

From a point P on a level ground , the angle of elevation to the top of the tower is 30° . If the tower is 100m high , the distance of point P from the foot of the tower is ( Take √3 = 1.73)

Accepted Answer

173 m

Question 11

A tower is broken at a point P above the ground. The top of the tower makes an angle 60° with the ground at Q. From another point R on the opposite sideof Q angle of elevation of point P is 30°. If QR = 180 m, then what is the total height (in metre…

Accepted Answer

45(√3+2)

Question 12

On level ground, there stands a vertical tower with a flagpole fixed on its top. From a point on the ground 9 m away from the foot of the tower, the angles of elevation of the top and the bottom of the flagpole are 60° and 30° respectively. What is …

Accepted Answer

Introduction / Context: This problem is a standard height and distance question based on right angled triangles and trigonometric ratios. A vertical tower has a flagpole fixed on its top, and an observer stands on level ground at a known horizontal distance from the foot of the tower. From that point, the observer measures two angles of elevation: one to the bottom of the flagpole (the top of the tower itself), and the other to the top of the flagpole. Using these two angles and the known horizontal distance, we can set up two tangent relations and then find the unknown height of the flagpole alone, without needing to know the total height of the tower in advance. Given Data / Assumptions: - The tower is vertical and stands on level ground. - A flagpole is fixed vertically on the top of the tower. - The observer is 9 m away from the foot of the tower along level ground. - Angle of elevation to the bottom of the flagpole (top of tower) is 30°. - Angle of elevation to the top of the flagpole is 60°. - We assume standard trigonometric values: tan 30° = 1 / √3 and tan 60° = √3. Concept / Approach: The key idea is to treat the situation as two right angled triangles that share the same horizontal base (9 m) but have different vertical heights. Let the height of the tower be H and the height of the flagpole be h. Then the total height from ground to the top of the flagpole is H + h. The angle of elevation to the top of the tower uses H in the tangent ratio, while the angle of elevation to the top of the flagpole uses H + h. By writing two tangent equations in terms of H and H + h and then subtracting, we can eliminate H and solve directly for h, the height of the flagpole. Step-by-Step Solution: Step 1: Let H be the height of the tower (up to the bottom of the flagpole) and h be the height of the flagpole itself. Step 2: From the point on the ground, the angle of elevation to the bottom of the flagpole is 30°, so tan 30° = H / 9. Step 3: Using tan 30° = 1 / √3, we get H / 9 = 1 / √3, so H = 9 / √3 = 3√3. Step 4: The angle of elevation to the top of the flagpole is 60°, so tan 60° = (H + h) / 9. Step 5: Using tan 60° = √3, we have (H + h) / 9 = √3, so H + h = 9√3. Step 6: Substitute H = 3√3 into H + h = 9√3 to get 3√3 + h = 9√3. Step 7: Rearranging gives h = 9√3 - 3√3 = 6√3. Step 8: Therefore the height of the flagpole is 6√3 m, which is approximately 10.39 m if we need a decimal value. Verification / Alternative check: We can verify the result by checking both triangles numerically. If H = 3√3 m and h = 6√3 m, then H + h = 9√3 m. Using the distance 9 m, tan 30° = H / 9 = (3√3) / 9 = √3 / 3, which matches the standard value of tan 30°. For the top of the flagpole, tan 60° = (H + h) / 9 = (9√3) / 9 = √3, which matches tan 60°. Since both trigonometric relations are satisfied exactly, the height 6√3 m is consistent and correct. Why Other Options Are Wrong: Option A (5√3 m) gives a total height H + h = 3√3 + 5√3 = 8√3 m, which would not produce tan 60° = √3 with a base of 9 m. Option C (6√2 m) does not match the standard tangent values for 30° and 60° when substituted back into the equations. Option D (6√5 m) similarly fails the tangent checks and does not satisfy both angle conditions simultaneously. Only 6√3 m preserves the exact trigonometric relationships for both given angles from the same observation point. Common Pitfalls: A common mistake is to treat the two given angles as if they belonged to two different base distances instead of the same 9 m base. Another frequent error is mixing up which angle corresponds to which height (tower top versus flagpole top), or assuming that the tower height is given directly. Students may also incorrectly use sine instead of tangent for heights over horizontal distances in such problems. Careful diagram drawing, correct identification of opposite and adjacent sides, and writing separate equations for each angle help avoid these errors. Final Answer: The height of the flagpole is 6√3 m, which corresponds to option B.

Question 13

Two ships are sailing in the sea on opposite sides of a vertical lighthouse. From the decks of the two ships, the angles of elevation of the top of the lighthouse are observed to be 30° and 45° respectively. If the lighthouse is 100 m high (take √3 …

Accepted Answer

Introduction / Context: This question involves two ships located on opposite sides of a vertical lighthouse at sea. Observers on each ship measure the angle of elevation of the top of the lighthouse. One ship observes an angle of 30°, while the other observes an angle of 45°. Since the height of the lighthouse is known, we can treat each observation as a separate right angled triangle sharing the same vertical height but having different horizontal distances from the lighthouse. By using basic trigonometric ratios, we can find the distance of each ship from the lighthouse and then add these distances to obtain the distance between the two ships. Given Data / Assumptions: - The lighthouse is vertical and 100 m high. - Two ships are on opposite sides of the lighthouse. - Angle of elevation from the first ship is 30°. - Angle of elevation from the second ship is 45°. - Sea level is considered to be a horizontal plane. - We use tan 30° = 1 / √3 and tan 45° = 1. Concept / Approach: The height of the lighthouse is the same for both ships, so we form two right angled triangles that share the same vertical side (100 m) but have different horizontal bases. In each triangle, the tangent of the angle of elevation is equal to the height divided by the horizontal distance from the lighthouse. For the ship with a 45° angle, the calculation is particularly simple because tan 45° = 1, so its distance from the lighthouse equals the height. For the ship with a 30° angle, we use tan 30° = 1 / √3 to find a larger horizontal distance. The total distance between the ships is the sum of these two horizontal distances because they lie on opposite sides of the lighthouse. Step-by-Step Solution: Step 1: Let d1 be the distance of the ship where the angle of elevation is 30°. Step 2: Let d2 be the distance of the ship where the angle of elevation is 45°. Step 3: For the 45° angle, tan 45° = 100 / d2. Since tan 45° = 1, we get 1 = 100 / d2, so d2 = 100 m. Step 4: For the 30° angle, tan 30° = 100 / d1. Using tan 30° = 1 / √3, we have 1 / √3 = 100 / d1, so d1 = 100√3 m. Step 5: The distance between the two ships is d1 + d2 = 100√3 + 100. Step 6: If we use √3 ≈ 1.73, then 100√3 ≈ 173 m, so d1 + d2 ≈ 173 + 100 = 273 m. Step 7: Therefore, the distance between the two ships is 273 m. Verification / Alternative check: We can verify the result numerically. If one ship is 100 m from the lighthouse, then the angle of elevation is tan⁻¹(100 / 100) = tan⁻¹(1) = 45°, which is correct. If the other ship is at 100√3 m, then the angle of elevation is tan⁻¹(100 / (100√3)) = tan⁻¹(1 / √3) = 30°, which also matches the given data. Since both angles are satisfied with these distances and the sum of the distances matches one of the answer options, the result is consistent. Why Other Options Are Wrong: Option A (173 m) corresponds roughly to 100√3 alone and ignores the distance of the second ship. Option B (200 m) underestimates the total distance, as the ships are clearly more than 100 m apart. Option D (300 m) is larger than the computed value and does not follow from tan 30° and tan 45° calculations. Only 273 m correctly represents d1 + d2 when both tangent relations are applied correctly. Common Pitfalls: Students sometimes forget that the ships are on opposite sides of the lighthouse and mistakenly subtract the distances instead of adding them. Another common mistake is using sine or cosine instead of tangent when the height and horizontal distance are involved. Mixing up which angle corresponds to which distance or using an incorrect value for √3 can also lead to errors. Being careful with which sides of the right triangle correspond to opposite and adjacent and remembering that distance between the ships is the sum of their distances from the lighthouse avoids these issues. Final Answer: The distance between the two ships is 273 m, which corresponds to option C.

Question 14

A man stands on the bank of a river and observes that the angle of elevation of the top of a tree on the opposite bank is 60°. When he walks 36 m directly away from the river bank, the angle of elevation of the top of the same tree becomes 30°. What…

Accepted Answer

Introduction / Context: This question is a classic example of using trigonometric ratios with two different observation points. A man observes the top of a tree across a river from two positions along a straight line perpendicular to the river bank. By measuring how the angle of elevation changes when he moves a known distance farther away, we can determine the unknown width of the river without directly measuring it. The tree is assumed to be vertical, and the ground on the man's side of the river is taken as level. Given Data / Assumptions: - The tree stands vertically on the opposite bank of the river. - The man initially stands at the river bank. - Angle of elevation of the top of the tree at the bank is 60°. - He then walks 36 m away from the bank in a straight line perpendicular to the river. - From the new point, the angle of elevation of the top of the tree is 30°. - The river banks are assumed parallel and the ground is level. Concept / Approach: Let the breadth of the river be w metres and the height of the tree be h metres. When the man is at the river bank, the horizontal distance from him to the tree is w. When he moves 36 m away, the distance becomes w + 36. The angle of elevation in each case gives us a tangent relation: tan 60° = h / w and tan 30° = h / (w + 36). Because both expressions equal the same tree height h, we can equate them and solve for w. This method avoids the need to calculate h explicitly until after we have found w, which is the required breadth of the river. Step-by-Step Solution: Step 1: Let w be the width of the river and h be the height of the tree. Step 2: At the river bank, the angle of elevation is 60°, so tan 60° = h / w. Step 3: Using tan 60° = √3, we get h / w = √3, so h = w√3. Step 4: After walking 36 m away from the bank, the horizontal distance becomes w + 36, and the angle of elevation is 30°. Step 5: For this position, tan 30° = h / (w + 36). Using tan 30° = 1 / √3, we have 1 / √3 = h / (w + 36). Step 6: Substitute h = w√3 into the second equation to get 1 / √3 = (w√3) / (w + 36). Step 7: Multiply both sides by (w + 36)√3 to get w + 36 = 3w. Step 8: Rearranging gives 3w - w = 36, so 2w = 36 and w = 18. Step 9: Therefore the breadth of the river is 18 m. Verification / Alternative check: We can verify by computing the height h. From h = w√3 with w = 18, we get h = 18√3 m. At the bank, tan 60° = h / w = (18√3) / 18 = √3, which is correct. At the new point, the distance is w + 36 = 18 + 36 = 54 m. The tangent of the new angle is h / (w + 36) = (18√3) / 54 = √3 / 3 = 1 / √3, which is the correct value for tan 30°. Both angles match the problem statement, confirming that w = 18 m is correct. Why Other Options Are Wrong: Option A (15 m) and option C (16 m) do not satisfy both tangent equations when substituted, so the angles will not come out as exactly 60° and 30°. Option D (11 m) is even smaller and clearly inconsistent with the geometry of the movement of 36 m away from the bank. Only 18 m produces a consistent pair of distances and heights that give the given angles of elevation. Common Pitfalls: One common mistake is to confuse which distance corresponds to which angle, for example taking w + 36 for the 60° case and w for the 30° case. Another error is to subtract 36 from w in the second position instead of adding, which does not match the described motion. Some students also incorrectly try to solve for h first in one equation and plug into the other without careful algebra, leading to sign errors. Drawing a clear diagram with both positions labelled and writing separate tangent equations helps avoid these issues. Final Answer: The breadth of the river is 18 m, which corresponds to option B.

Question 15

A kite is flying in the sky. The length of the string between a point on the level ground and the kite is 420 m. The angle of elevation that the string makes with the ground is 30°. Assuming that the string is straight and taut with no slack, what i…

Accepted Answer

Introduction / Context: This problem involves a simple right angled triangle formed by the kite, the point on the ground directly below the kite, and the point where the person is holding the string. The length of the string and the angle it makes with the ground are given. Assuming the string is straight and taut, the string itself becomes the hypotenuse of the right triangle, while the height of the kite is the side opposite the given angle of elevation. Using basic trigonometry, we can directly compute the vertical height of the kite above the ground. Given Data / Assumptions: - Length of the kite string (hypotenuse) is 420 m. - Angle of elevation of the string with the horizontal ground is 30°. - The string is taut and straight, so there is no slack or sag. - The ground between the observer and the point directly below the kite is level. Concept / Approach: We consider a right angled triangle where the hypotenuse is the string of length 420 m, the side adjacent to the 30° angle is the horizontal distance on the ground, and the side opposite the 30° angle is the height of the kite. The appropriate trigonometric ratio is sine, since sin of an angle is equal to the length of the opposite side divided by the length of the hypotenuse. Thus, if the height is h, then sin 30° = h / 420. Since sin 30° has a simple standard value, this calculation is straightforward. Step-by-Step Solution: Step 1: Let h be the height of the kite above the ground. Step 2: The string of length 420 m is the hypotenuse of the right angled triangle. Step 3: Using the sine ratio, sin 30° = opposite / hypotenuse = h / 420. Step 4: The standard value is sin 30° = 1 / 2. Step 5: Substitute sin 30° into the equation: 1 / 2 = h / 420. Step 6: Multiply both sides by 420 to get h = 420 * (1 / 2) = 210. Step 7: Therefore, the height of the kite is 210 m. Verification / Alternative check: To verify, we can also check using the cosine ratio to find the horizontal distance and see if it forms a consistent triangle. If h = 210 m and the hypotenuse is 420 m, then the horizontal distance d satisfies d^2 + h^2 = 420^2. Using cos 30° = √3 / 2, we get d = 420 * (√3 / 2) = 210√3. Now check: d^2 + h^2 = (210√3)^2 + 210^2 = 210^2 * 3 + 210^2 = 210^2 * 4 = (210 * 2)^2 = 420^2, which confirms that the triangle is consistent and that h = 210 m is correct. Why Other Options Are Wrong: Option B (140√3) would be larger than 210 and does not match sin 30° = 1 / 2 when used with a 420 m hypotenuse. Option C (210√3) is even larger and corresponds to the horizontal distance, not the vertical height. Option D (150) is less than half of 420 and does not satisfy sin 30° = 1 / 2. Only 210 m makes the sine ratio exact and consistent with the given angle and hypotenuse length. Common Pitfalls: Students sometimes confuse sine and cosine and mistakenly compute h using cos 30° instead of sin 30°, which would actually give the horizontal distance rather than the height. Another common error is treating 420 m as the horizontal distance instead of the hypotenuse. Finally, using approximate decimal values for sine and cosine without understanding which side is opposite and which is adjacent can lead to incorrect results. Remembering that sin θ = opposite / hypotenuse and cos θ = adjacent / hypotenuse and clearly labelling the triangle helps avoid these mistakes. Final Answer: The height of the kite above the ground is 210 m, which corresponds to option A.

Question 16

Two vertical trees stand on opposite sides of a straight road, and the distance between the two trees is 400 metres. There is a point on the road between them from which the angles of depression to that point from the tops of the trees are 45° and 6…

Accepted Answer

Introduction / Context: This problem combines the concept of angles of depression with distances along a straight road. Two trees of different heights are located on opposite sides of a road, and a point on the road between them is observed from the top of each tree. The angles of depression to that same point are different, and the height of one tree is known. The total distance between the trees along the road is also given. We need to use right angled triangles and tangent ratios to compute the height of the second tree. Given Data / Assumptions: - Two trees stand vertically on opposite sides of a straight road. - The horizontal distance between the two trees is 400 m. - There is a point P on the road between the trees. - Angle of depression from the top of the first tree to point P is 45°. - Angle of depression from the top of the second tree to point P is 60°. - The height of the tree corresponding to the 45° angle is 200 m. - The ground and the road are level. Concept / Approach: An angle of depression from the top of a tree to a point on the ground is equal to the angle of elevation from that point to the top of the tree. Therefore, from point P on the road, the angles of elevation to the tops of the trees are 45° and 60°. For the tree of height 200 m with a 45° angle of elevation, the horizontal distance from P to that tree is easy to compute because tan 45° = 1. Once we know that distance, we can find the distance from P to the second tree by subtracting from the total distance of 400 m. With that horizontal distance and the 60° angle of elevation, we can then use tan 60° to find the height of the second tree. Step-by-Step Solution: Step 1: Let tree A be the one with height 200 m and angle of depression 45° to point P. Step 2: Let tree B be the unknown height tree with angle of depression 60° to the same point P. Step 3: From point P, the angle of elevation to the top of tree A is 45°, so tan 45° = 200 / dA, where dA is the horizontal distance from P to tree A. Step 4: Since tan 45° = 1, we have 1 = 200 / dA, so dA = 200 m. Step 5: The total distance between the trees is 400 m, so the distance from P to tree B is dB = 400 - dA = 400 - 200 = 200 m. Step 6: From point P, the angle of elevation to the top of tree B is 60°, so tan 60° = hB / dB, where hB is the height of tree B. Step 7: Using tan 60° = √3, we get √3 = hB / 200, so hB = 200√3. Step 8: Therefore, the height of the second tree is 200√3 m. Verification / Alternative check: We can check the logic by considering the geometry. If tree A has height 200 m and the horizontal distance to P is also 200 m, then tan of the angle of elevation is 200 / 200 = 1, which corresponds to 45°. For tree B, we use the same horizontal distance 200 m because P is exactly midway between the two trees. With hB = 200√3, we get tan of the angle of elevation as (200√3) / 200 = √3, which corresponds to 60°. Both angles match the given angles of depression, confirming that the distances and heights are consistent. Why Other Options Are Wrong: Option A (200 m) would imply both trees have the same height, which would give the same angle of depression from each tree, contradicting the statement that one angle is 45° and the other is 60°. Option C (100√3 m) is smaller than 200√3 and would not produce a 60° angle with the same horizontal distance. Option D (250 m) does not relate neatly to tan 60° or the given distance of 200 m from P, and the resulting tangent would not be √3. Only 200√3 m is consistent with the correct tangent ratio and the geometry of the problem. Common Pitfalls: Some learners mistakenly assume that the point on the road is at one end rather than between the trees, leading to incorrect distance calculations. Others forget that angle of depression from the tree equals the angle of elevation from the point on the road and end up misplacing the angles in their diagrams. Another typical error is to use sine or cosine instead of tangent when relating heights to horizontal distances. Drawing a clear diagram and using tan θ = opposite / adjacent for each tree avoids these common mistakes. Final Answer: The height of the other tree is 200√3 m, which corresponds to option B.

Question 17

A Navy captain in a boat is moving away from the foot of a vertical lighthouse at a constant speed of 4(√3 − 1) m/s. He observes that it takes 1 minute for the angle of elevation of the top of the lighthouse to change from 60° to 45°. What is the he…

Accepted Answer

Introduction / Context: This problem combines motion in a straight line with trigonometric height and distance concepts. A Navy captain in a boat moves directly away from a lighthouse at a known constant speed. As he moves, the angle of elevation of the top of the lighthouse changes from 60° to 45° in a given time interval. Because the height of the lighthouse does not change, we can express the captain's varying horizontal distances from the lighthouse in terms of the height and the observed angles. The change in horizontal distance over the known time must match the distance travelled using speed multiplied by time, which allows us to solve for the unknown height of the lighthouse. Given Data / Assumptions: - The lighthouse is vertical and stands on level sea or ground. - The captain moves in a straight line directly away from the lighthouse. - Speed of the boat is 4(√3 − 1) m/s. - Time taken for the angle of elevation to change from 60° to 45° is 1 minute (60 seconds). - At the initial position, angle of elevation of the top of the lighthouse is 60°. - At the later position, angle of elevation is 45°. - We use tan 60° = √3 and tan 45° = 1. Concept / Approach: Let H be the height of the lighthouse. At the first observation point, the captain is at a horizontal distance x from the foot of the lighthouse, forming a right angled triangle with height H. At the second observation point, after moving further away, he is at a distance x2 from the foot. From the definition of tangent, tan 60° = H / x and tan 45° = H / x2. These give x = H / √3 and x2 = H. The captain travels from x to x2 in 60 seconds, so the increase in distance x2 − x must equal speed multiplied by time, that is 4(√3 − 1) * 60. Equating these leads to a simple equation in H that we can solve. Step-by-Step Solution: Step 1: Let H be the height of the lighthouse in metres. Step 2: At the first position, angle of elevation is 60°, so tan 60° = H / x, where x is the initial horizontal distance. Step 3: Since tan 60° = √3, we get √3 = H / x, so x = H / √3. Step 4: At the second position, the angle of elevation is 45°, so tan 45° = H / x2, where x2 is the new horizontal distance. Step 5: Because tan 45° = 1, we get 1 = H / x2, so x2 = H. Step 6: The captain travels from distance x to distance x2 in 60 seconds, so the increase in horizontal distance is x2 − x = H − H / √3. Step 7: His speed is 4(√3 − 1) m/s, so the distance travelled in 60 seconds is 4(√3 − 1) * 60 = 240(√3 − 1). Step 8: Equate the two expressions for the distance travelled: H − H / √3 = 240(√3 − 1). Step 9: Factor H on the left: H(1 − 1 / √3) = 240(√3 − 1). Step 10: Write 1 − 1 / √3 as (√3 − 1) / √3. Then H * (√3 − 1) / √3 = 240(√3 − 1). Step 11: Cancel the common factor (√3 − 1) from both sides, giving H / √3 = 240. Step 12: Multiply both sides by √3 to get H = 240√3. Step 13: Thus, the height of the lighthouse is 240√3 m. Verification / Alternative check: We can verify the result numerically. Using √3 ≈ 1.732, H ≈ 240 * 1.732 ≈ 415.7 m. Then x = H / √3 ≈ 415.7 / 1.732 ≈ 240 m and x2 = H ≈ 415.7 m. The distance travelled is x2 − x ≈ 415.7 − 240 ≈ 175.7 m. Now compute the distance using speed and time: speed = 4(√3 − 1) ≈ 4(1.732 − 1) ≈ 4 * 0.732 ≈ 2.928 m/s. In 60 seconds, distance ≈ 2.928 * 60 ≈ 175.7 m, which matches the geometry, confirming that H = 240√3 m is consistent. Why Other Options Are Wrong: Option B (480(√3 − 1)) represents the total distance travelled, not the height of the lighthouse. Option C (360√3) and option D (280√2) do not satisfy the relationship between the two tangent equations and the distance travelled when substituted into the same reasoning. Only 240√3 m simultaneously satisfies both the trigonometric conditions and the motion constraint given by the speed and time. Common Pitfalls: Students often confuse the roles of x and x2, sometimes treating both as equal or forgetting that the second distance must be greater. Another common error is to use the wrong trigonometric ratio (such as sine or cosine) instead of tangent when relating height to horizontal distance. Algebraic mistakes can also occur when simplifying 1 − 1 / √3. Carefully expressing the problem in terms of tan θ = height / base at each point and explicitly equating the change in distance to speed multiplied by time helps prevent these issues. Final Answer: The height of the lighthouse is 240√3 m, which corresponds to option A.

Question 18

From the top of a vertical tower 60 m high, the angles of depression of the top and the bottom of a vertical pole are observed to be 45° and 60° respectively. If the pole and the tower stand on the same horizontal plane, what is the height of the po…

Accepted Answer

Introduction / Context: This height and distance question involves angles of depression from the top of a tower to both the top and the bottom of a shorter pole standing some distance away on the same level ground. Angles of depression from a horizontal line of sight are equal to angles of elevation from the point on the ground to the observer. By converting these angles of depression into angles of elevation and using tangent ratios, we can relate the heights and the horizontal distance between the tower and the pole. This allows us to determine the unknown height of the pole using simple trigonometry. Given Data / Assumptions: - The tower is vertical and 60 m high. - A vertical pole stands on the same horizontal plane as the tower. - From the top of the tower, the angle of depression to the top of the pole is 45°. - From the top of the tower, the angle of depression to the bottom (foot) of the pole is 60°. - The horizontal distance between the tower and the pole is unknown. - Ground is level, and both structures are perpendicular to the ground. Concept / Approach: Let h be the height of the pole and d be the horizontal distance between the tower and the pole. The line of sight from the top of the tower to the top of the pole forms a right angled triangle whose vertical side is the difference in heights (60 − h) and whose base is d. The angle at the tower top with respect to the horizontal is 45°. Similarly, the line of sight from the top of the tower to the bottom of the pole forms another right angled triangle with vertical side 60 and base d and an angle of 60°. Using tan 45° and tan 60° on these two triangles gives us two expressions involving d and h. Solving these simultaneously allows us to find the exact height of the pole. Step-by-Step Solution: Step 1: Let h be the height of the pole in metres, and let d be the horizontal distance between the tower and the pole. Step 2: Consider the angle of depression of 60° from the top of the tower to the bottom of the pole. The vertical opposite side is the full height of the tower, 60 m, and the horizontal adjacent side is d. Step 3: Using the tangent ratio, tan 60° = 60 / d. Since tan 60° = √3, we get √3 = 60 / d, so d = 60 / √3. Step 4: Simplify d: d = 60 / √3 = (60√3) / 3 = 20√3. Step 5: Now consider the angle of depression of 45° from the top of the tower to the top of the pole. The vertical difference in height between the top of the tower and the top of the pole is 60 − h, and the horizontal distance is still d. Step 6: Using tan 45° = (60 − h) / d and tan 45° = 1, we have 1 = (60 − h) / d, so 60 − h = d. Step 7: Substitute d = 20√3 into 60 − h = d to obtain 60 − h = 20√3. Step 8: Rearranging gives h = 60 − 20√3. Step 9: Factor out 20 from the expression: h = 20(3 − √3). Step 10: Therefore, the exact height of the pole is 20(3 − √3) m. Verification / Alternative check: We can check this result numerically. Using √3 ≈ 1.732, h ≈ 20(3 − 1.732) = 20 * 1.268 ≈ 25.36 m. Then the difference in heights is 60 − h ≈ 34.64 m and the distance d is 20√3 ≈ 34.64 m. For the line of sight to the top of the pole, tan of the angle is (60 − h) / d ≈ 34.64 / 34.64 = 1, which corresponds to 45°. For the line of sight to the bottom of the pole, tan of the angle is 60 / d ≈ 60 / 34.64 ≈ 1.732, which corresponds to √3 and hence to 60°. Both angles match the given information, confirming that the height 20(3 − √3) m is correct. Why Other Options Are Wrong: Option B (20(3 + √3) m) is larger than 60 m when evaluated, which is impossible for the pole height since the tower height is only 60 m. Option C (20(√3 + 1) m) gives a height that, when substituted into 60 − h, does not equal the computed horizontal distance 20√3, so it fails the tan 45° relation. Option D (20(3√3) m) is much larger than 60 m and is clearly inconsistent with the given tower height. Only 20(3 − √3) m satisfies both tangent equations exactly and is less than the tower height, as expected for a shorter pole. Common Pitfalls: A typical mistake is to misinterpret angles of depression as angles of elevation without correctly reflecting them across the horizontal line. This can lead to swapping the roles of vertical and horizontal distances or incorrectly identifying which angle corresponds to which segment. Another error is forgetting that both lines of sight use the same horizontal distance d, causing students to introduce multiple distances unnecessarily. Algebraic errors can also occur when simplifying d = 60 / √3 or when solving for h in 60 − h = d. Careful diagram drawing and step-by-step use of tan θ = opposite / adjacent for each angle help avoid these problems. Final Answer: The height of the pole is 20(3 − √3) m, which corresponds to option A.

Question 19

If the elevation of the Sun changes from 30° to 60°, what is the difference between the lengths of the shadows of a vertical pole 15 m high?

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Introduction / Context: This problem is a standard height and distance question based on the shadow of a vertical pole when the elevation of the Sun changes. The main concept tested here is the use of trigonometric ratios, especially the tangent function, to relate the height of an object with the length of its shadow on level ground. You are asked to compare the shadow lengths at two different angles of elevation and to find the difference between those lengths for a pole of fixed height 15 m. Given Data / Assumptions: The height of the vertical pole is 15 m. First position: elevation of the Sun is 30 degrees. Second position: elevation of the Sun is 60 degrees. The ground is assumed to be perfectly horizontal and the pole is perfectly vertical. We use tan θ = opposite / adjacent, where opposite is the height of the pole and adjacent is the length of the shadow. Concept / Approach: In a right angled triangle formed by the pole, its shadow and the line of sight to the top, the tangent of the angle of elevation equals height / shadow length. So, shadow length = height / tan θ. We calculate the shadow at 30 degrees and at 60 degrees, then subtract the smaller from the larger to obtain the required difference. The key trigonometric values used are tan 30 degrees = 1 / sqrt(3) and tan 60 degrees = sqrt(3). Step-by-Step Solution: Let the height of the pole be h = 15 m. When the Sun's elevation is 30 degrees, tan 30 = h / shadow_30. So shadow_30 = h / tan 30 = 15 / (1 / sqrt(3)) = 15 * sqrt(3) m. When the Sun's elevation is 60 degrees, tan 60 = h / shadow_60. So shadow_60 = h / tan 60 = 15 / sqrt(3) = 5 * sqrt(3) m. Difference in shadow lengths = shadow_30 - shadow_60 = 15√3 - 5√3 = 10√3 m. Verification / Alternative check: We know that as the Sun rises higher in the sky (angle increases), the shadow becomes shorter. At 30 degrees the shadow should be longer than at 60 degrees. Our calculation gives 15√3 m at 30 degrees and 5√3 m at 60 degrees, so the first shadow is indeed longer. The difference of 10√3 m is positive and dimensionally consistent (in metres), which confirms the internal logic of the solution. Why Other Options Are Wrong: 7.5 m: This ignores the presence of sqrt(3) in the tangent values and is numerically too small. 15 m: This wrongly assumes a direct difference in height instead of using trigonometric ratios. 5√3 m: This corresponds roughly to the shorter shadow alone, not to the difference of the two shadows. 20 m: This is an arbitrary value with no correct trigonometric basis for the given angles. Common Pitfalls: Students often mix up tan 30 degrees and tan 60 degrees, or use sin and cos instead of tan for shadow problems. Another common mistake is to subtract in the wrong order, taking shorter minus longer and getting a negative length, which is not meaningful here. Forgetting that shadow length = height / tan θ, not height * tan θ, is also a frequent source of error. Final Answer: The difference between the lengths of the shadows of the 15 m high pole when the Sun's elevation changes from 30 degrees to 60 degrees is 10√3 m.

Question 20

At a point 129 m from the foot of a cliff on level ground, the angle of elevation of the top of the cliff is 30°. What is the height of the cliff?

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Introduction / Context: This question checks your understanding of basic trigonometry in right angled triangles, specifically the use of the tangent ratio to find the height of a vertical object such as a cliff. You are given the horizontal distance from the observer to the foot of the cliff and the angle of elevation to the top. From this information, you need to determine the height of the cliff using a simple relation involving tan 30 degrees. Given Data / Assumptions: Horizontal distance from the observer to the foot of the cliff = 129 m. Angle of elevation to the top of the cliff = 30 degrees. The ground between the observer and the foot of the cliff is level. The cliff is vertical, so the triangle formed is right angled at the foot. Standard trigonometric value: tan 30 degrees = 1 / sqrt(3). Concept / Approach: In a right triangle, tan θ = opposite / adjacent. Here, the opposite side is the height of the cliff and the adjacent side is the horizontal distance from the observer to the foot of the cliff (129 m). Rearranging the formula gives height = distance * tan θ. Substituting θ = 30 degrees and tan 30 = 1 / sqrt(3) will provide the exact height in terms of sqrt(3). Step-by-Step Solution: Let the height of the cliff be H. We know that tan 30 = H / 129. Substitute tan 30 = 1 / sqrt(3): 1 / sqrt(3) = H / 129. So H = 129 * (1 / sqrt(3)) = 129 / sqrt(3). Rationalise: 129 / sqrt(3) = (129 * sqrt(3)) / 3 = 43 * sqrt(3). Therefore, the height of the cliff is 43√3 m. Verification / Alternative check: We know that tan 30 degrees is small (about 0.577). Multiplying 129 by approximately 0.577 gives about 74.5 m. Now 43√3 is roughly 43 * 1.732 which is around 74.5 m as well. So the computed expression 43√3 m is numerically consistent with the approximate calculation and confirms that the result is reasonable. Why Other Options Are Wrong: 50√3 m: This is larger than what tan 30 degrees would give for a distance of 129 m. 45√3 m: Slightly too large when compared with the accurate product 129 * tan 30. 47√3 m: Also overestimates the height based on the given angle and distance. 40√3 m: Underestimates the height and does not satisfy H = 129 * tan 30. Common Pitfalls: A common error is to interchange the roles of opposite and adjacent sides, which leads to using H = 129 / tan 30 instead of H = 129 * tan 30. Another frequent mistake is to use approximate values of tan incorrectly or to confuse tan 30 with tan 60. Always remember the exact standard values: tan 30 = 1 / sqrt(3), tan 45 = 1, tan 60 = sqrt(3). Final Answer: The height of the cliff is 43√3 m.

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