Question 1

Right triangle at a house corner: At a point 15 m from the base of a 15 m high house on level ground, what is the angle of elevation of the top of the house?

Accepted Answer

Introduction / Context: Equal opposite and adjacent sides in a right triangle imply a 45° angle of elevation by the tangent definition. Given Data / Assumptions: Height = 15 m; horizontal distance = 15 m; level ground; vertical wall. Concept / Approach: tan θ = opposite/adjacent = 15/15 = 1 → θ = 45°. Step-by-Step Solution: tan θ = 1 → θ = 45°. Verification / Alternative check: Pythagoras gives hypotenuse 15√2; this corresponds to an isosceles right triangle. Why Other Options Are Wrong: 30° or 60° would require unequal legs; 90° is impossible for a finite distance. Common Pitfalls: Using sine instead of tangent; confusing angle at the top vs at the observer. Final Answer: 45°

Question 2

Two-level observation: elevation and depression: A vertical tower subtends an angle of 30° at a ground-level point P. From a second point directly above P by h metres (same vertical line), the angle of depression to the foot of the tower is 60°. Fin…

Accepted Answer

Introduction / Context: One observation is an elevation from the ground; the other is a depression from a point vertically above it. The horizontal distance remains the same for both, allowing a direct relation to h via tangent of 60°. Given Data / Assumptions: At P (ground), angle of elevation to top = 30°. At Q (directly above P by h), angle of depression to foot = 60°. Horizontal distance from P (and Q) to tower base = x (unknown). Concept / Approach: From Q to the tower foot: tan 60° = vertical drop / horizontal = h / x → x = h / √3 = h cot 60°. The first observation would then set the tower height, but the question only asks x. Step-by-Step Solution: tan 60° = h / x → x = h / √3 = h cot 60°.Consistency check (optional): If tower height is T, tan 30° = T / x → T = x / √3 = h / 3, which is positive and consistent. Verification / Alternative check: Units: x and h are lengths; cot 60° is dimensionless; expression is dimensionally sound. Why Other Options Are Wrong: h cot 30° equals h√3, not supported by tan 60° relation; the “squared” looking options are not meaningful here. Common Pitfalls: Using elevation (30°) to compute x directly; mixing up which angle applies to which vertical difference; confusing depression with elevation. Final Answer: h cot 60°

Question 3

Height & Distance – Sun’s altitude from shadow ratio A vertical pole casts a shadow on level ground. If the length of the shadow is √3 times the height of the pole, what is the angle of elevation of the Sun (measured from the horizontal)?

Accepted Answer

Introduction / Context: Problems that connect a pole’s height and its shadow use right-triangle trigonometry on level ground. The angle of elevation θ of the Sun satisfies tan θ = (opposite)/(adjacent) = height/shadow. Here, the shadow is a known multiple of the height, so the triangle's ratio is fixed. Given Data / Assumptions: Pole is vertical; ground is horizontal. Shadow length = √3 × (pole height). Angle of elevation θ is acute and measured from the horizontal. Concept / Approach: Let height = h and shadow = √3 h. Then tan θ = h/(√3 h) = 1/√3. The standard acute angle whose tangent is 1/√3 is 30°. Step-by-Step Solution: Let height = h > 0Shadow = √3 htan θ = height / shadow = h / (√3 h) = 1/√3Hence θ = 30° Verification / Alternative check: For θ = 30°, the 30°–60°–90° triangle has opposite:adjacent = 1:√3, matching the given shadow-to-height ratio. Why Other Options Are Wrong: 60° gives tan = √3, not 1/√3. 90° is vertical Sun (shadow 0). 45° gives tan = 1, which requires equal height and shadow, not √3 times. Common Pitfalls: Inverting the ratio (taking shadow/height instead of height/shadow), or confusing 30° and 60° special-angle tangents. Final Answer: 30°

Question 4

Height & Distance – Wire length from angle Two poles are 20 m and 14 m high. Their tops are connected by a straight wire that makes a 30° angle with the horizontal. What is the length of the wire?

Accepted Answer

Introduction / Context: When a straight wire spans between unequal heights, its inclination with the horizontal lets us relate the vertical difference to the wire’s length via basic trigonometry. The scenario forms a right triangle whose “opposite” is the height difference and whose hypotenuse is the wire. Given Data / Assumptions: Taller pole = 20 m; shorter pole = 14 m. Wire joins their tops; inclination to horizontal = 30°. Ground is level; wire is taut and straight. Concept / Approach: Let L be the wire length. Vertical rise between ends is Δh = 20 − 14 = 6 m. With angle θ to the horizontal, sin θ = (vertical rise)/L = Δh/L. Hence L = Δh / sin θ. Step-by-Step Solution: Δh = 6 msin 30° = 1/2L = Δh / sin 30° = 6 / (1/2) = 12 m Verification / Alternative check: Using components: vertical = L * sin 30° = L/2. Set L/2 = 6 ⇒ L = 12, consistent. Why Other Options Are Wrong: 10 m and 8 m correspond to sin values inconsistent with 30° for a 6 m rise. Common Pitfalls: Accidentally using tan instead of sin (tan ties vertical to horizontal span, not to hypotenuse), or mixing which pole is taller. Final Answer: 12 m

Question 5

Height & Distance – Finding pole height from two elevation observations The top of a 15 m tower subtends a 60° angle at the bottom of a nearby electric pole, and a 30° angle at the top of that pole. What is the height of the electric pole?

Accepted Answer

Introduction / Context: Two angles of elevation to the same top point (here, the top of a 15 m tower) taken from two different heights (pole bottom and pole top) allow us to determine the unknown pole height using tan relations along a common horizontal line. Given Data / Assumptions: Tower height = 15 m. Angle from pole bottom to tower top = 60°. Angle from pole top to tower top = 30°. Let horizontal separation between tower base and pole base be x > 0. Let pole height be h. Concept / Approach: From pole bottom: tan 60° = 15/x ⇒ x = 15/√3 = 5√3. From the pole top, vertical difference to tower top is (15 − h), and tan 30° = (15 − h)/x = 1/√3. Solve for h. Step-by-Step Solution: x = 5√3tan 30° = (15 − h)/x = 1/√3 ⇒ (15 − h) = x/√3 = 5h = 15 − 5 = 10 m Verification / Alternative check: Check the first observation: tan 60° = 15/(5√3) = √3, valid. Second: tan 30° = (15 − 10)/(5√3) = 5/(5√3) = 1/√3, valid. Why Other Options Are Wrong: 5 m, 8 m, 12 m give inconsistent tan values for one or both observations. Common Pitfalls: Swapping which angle corresponds to which observation point, or forgetting that the second observation sees only the difference in heights. Final Answer: 10 m

Question 6

Angle of depression + speed → time to reach directly beneath From a bridge 15 m above a river, the angle of depression to a boat is 30°. If the boat moves at 6 km/h on a straight path toward the bridge's vertical, how long (in seconds) will it take …

Accepted Answer

Introduction / Context: Angle-of-depression scenarios translate to angle-of-elevation from the boat to the bridge top. The right triangle’s vertical leg is fixed (bridge height), which gives the horizontal distance via tan. With the boat’s uniform speed, time = distance/speed after unit consistency. Given Data / Assumptions: Bridge height h = 15 m (vertical). Angle of depression = 30° ⇒ tan 30° = h/d. Boat speed = 6 km/h (straight toward the point below the bridge). Concept / Approach: Horizontal distance d = h / tan 30° = 15 / (1/√3) = 15√3 m. Convert 6 km/h to m/s, then compute time t = d / v. Step-by-Step Solution: d = 15√3 m6 km/h = 6000 m / 3600 s = 5/3 m/s ≈ 1.666… m/st = d / v = (15√3) / (5/3) = 9√3 s Verification / Alternative check: Compute numerically: √3 ≈ 1.732 ⇒ 9√3 ≈ 15.588 s. Distance 15√3 ≈ 25.98 m at 1.666… m/s takes ≈ 15.59 s, consistent. Why Other Options Are Wrong: 19/√3 s or 3√3 s do not satisfy d = v * t with the given geometry and speed. Common Pitfalls: Using sin or cos instead of tan for horizontal distance, or failing to convert km/h to m/s before computing time. Final Answer: 9 √3 second

Question 7

Height from angle of elevation at a known horizontal distance From a point on level ground 50 m from the tower base, the angle of elevation to the tower's top is 30°. What is the tower's height?

Accepted Answer

Introduction / Context: This is a direct tan relation in a right triangle. Angle of elevation θ at horizontal distance d gives height h = d * tan θ. Correctly evaluating tan 30° and simplifying radicals is all that is required. Given Data / Assumptions: Horizontal distance d = 50 m. Angle of elevation θ = 30°. Level ground, vertical tower. Concept / Approach: h = d * tan θ = 50 * tan 30° = 50 * (1/√3) = 50/√3 m. If rationalized, h = (50√3)/3 m ≈ 28.87 m. Check options for an exact match in form or value. Step-by-Step Solution: tan 30° = 1/√3h = 50 / √3 m = (50√3)/3 m Verification / Alternative check: Numerical value: (50√3)/3 ≈ 28.867 m. The provided options list only multiples of √3 like 50√3 m, 75√3 m, etc., none equivalent to 50/√3 m. Why Other Options Are Wrong: All listed values represent heights far larger than 50/√3 m or in a mismatched form. None equals 50/√3 m. Common Pitfalls: Using sin instead of tan, or flipping the ratio (taking h/50 = √3 instead of 1/√3). Also note the option formatting: “50 √3 m” is not the same as “50/√3 m.” Final Answer: None of these

Question 8

Evaluate an expression from r sinθ and r cosθ Given r sin θ = 1 and r cos θ = √3, find the value of ( √3 * tan θ + 1 ).

Accepted Answer

Introduction / Context: This is a trig identity/evaluation problem. Using the given r-scaled sine and cosine helps find tan θ directly by division, and the Pythagorean identity can verify consistency of r and θ. Given Data / Assumptions: r sin θ = 1 r cos θ = √3 We seek S = √3 * tan θ + 1 Concept / Approach: Compute tan θ = (r sin θ)/(r cos θ) = 1/√3. Then plug into S. Optionally, confirm r by squaring and summing to ensure a consistent setup. Step-by-Step Solution: tan θ = (r sin θ)/(r cos θ) = 1/√3S = √3 * tan θ + 1 = √3 * (1/√3) + 1 = 1 + 1 = 2Check: r^2 = (r sin θ)^2 + (r cos θ)^2 = 1 + 3 = 4 ⇒ r = 2 (consistent) Verification / Alternative check: If tan θ = 1/√3, then θ corresponds to 30° in the principal acute range; the expression evaluates to 2 as found. Why Other Options Are Wrong: Values like √3 or 1 occur if tan θ is misread as √3 or 0. Common Pitfalls: Dividing in the wrong order (cos/sin) or forgetting that √3 * (1/√3) simplifies to 1. Final Answer: 2

Question 9

Complementary elevations to two posts of heights in a 2:1 ratio Two vertical posts are x meters apart. One is twice as tall as the other. From the midpoint between their feet, the angles of elevation of their tops are complementary. Find the height …

Accepted Answer

Introduction / Context: When two elevation angles are complementary (α + β = 90°), the corresponding tangents satisfy tan α * tan β = 1. With symmetric horizontal distances (midpoint observer) and a 2:1 height ratio, the product-of-tangents condition yields the shorter height in terms of x. Given Data / Assumptions: Posts are x meters apart on level ground. Heights: shorter = h, taller = 2h. Observer is at midpoint; horizontal distance to each post = x/2. Angles of elevation to tops are complementary. Concept / Approach: tan α (to shorter) = h / (x/2) = 2h/x. tan β (to taller) = (2h) / (x/2) = 4h/x. Complementary ⇒ tan α * tan β = 1 ⇒ (2h/x)*(4h/x) = 1 ⇒ 8h^2 = x^2. Step-by-Step Solution: (2h/x) * (4h/x) = 1 ⇒ 8h^2/x^2 = 1h^2 = x^2 / 8h = x / (2√2) Verification / Alternative check: Plug h back to check tan α * tan β = 1 exactly; it holds for all x > 0. Why Other Options Are Wrong: x/4 or x/√2 do not satisfy the product condition; x√2 is dimensionally large vs. half-spacing geometry. Common Pitfalls: Using α + β = 90° without leveraging tan α * tan β = 1, or forgetting both horizontal legs are equal from the midpoint. Final Answer: x / (2√2)

Question 10

Two angles of depression to ships 200 m apart → lighthouse height From the top of a lighthouse, the angles of depression of two ships on the same straight line eastward are 45° and 30°. If the ships are 200 m apart along that line, find the height o…

Accepted Answer

Introduction / Context: Angles of depression equal angles of elevation from the horizontal at the ship. With both ships along the same ground line from the lighthouse foot (east), their horizontal distances differ by 200 m. Using tan relations for 45° and 30°, we solve for the common height. Given Data / Assumptions: Angles of depression: nearer ship 45°, farther ship 30° (or vice versa—but distances will sort themselves since 45° corresponds to the nearer one). Distance between ships along the line = 200 m. Let height be h and horizontal distances be d₁ (for 45°), d₂ (for 30°), with d₂ > d₁. Concept / Approach: tan 45° = h/d₁ ⇒ d₁ = h. tan 30° = h/d₂ ⇒ d₂ = h√3. The ships are 200 m apart, so d₂ − d₁ = 200 ⇒ h√3 − h = 200. Step-by-Step Solution: h(√3 − 1) = 200h = 200 / (√3 − 1) = 200(√3 + 1) / (3 − 1) = 100(√3 + 1)Numerically: √3 ≈ 1.732 ⇒ h ≈ 100(2.732) = 273.2 m ≈ 273 m Verification / Alternative check: Distances: d₁ ≈ 273.2 m; d₂ ≈ 473.2 m; difference ≈ 200 m, consistent. Why Other Options Are Wrong: 100 m and 173 m understate the geometry; 200 m is the separation, not the height. Common Pitfalls: Assigning the 30° to the nearer ship (which would invert distances), or forgetting to rationalize when comparing to rounded options. Final Answer: 273 m

Question 11

Tower + flagstaff seen from a point – formula recall At a point on level ground, a tower subtends angle θ at the eye. A flagstaff of length a fixed on the tower top subtends angle ϕ at the same point. What is the height of the tower alone (in terms …

Accepted Answer

Introduction / Context: This classic configuration stacks a flagstaff of known length a atop an unknown-height tower. From one observation point, the tower alone subtends angle θ, while the flagstaff alone subtends angle ϕ. Using angle-addition along a common line of sight yields a closed-form expression for the tower height in terms of a, θ, and ϕ. Given Data / Assumptions: Level ground; tower vertical; flagstaff vertical and on the tower's top. Angle subtended by tower at eye = θ. Angle subtended by flagstaff (length a) at eye = ϕ. All angles are acute; heights and distances are positive. Concept / Approach: Let H be the tower height, D be horizontal distance to the tower foot, and the line of sight to the tower top define angle θ. The flagstaff of length a adds an angular contribution that composes with θ to reach the line of sight of the flagstaff’s top. Resolving with right-triangle projections and the angle-addition identity produces the known result. Step-by-Step Solution (outline): tan θ = H / DLet the elevation to the flagstaff’s top be θ + δ; the flagstaff alone subtends ϕ.Trigonometric decomposition along the same baseline yields H = a * sin θ * cos ϕ / cos(θ + ϕ) Verification / Alternative check: Dimensional sanity: RHS has units of length (a multiplied by trigonometric ratios). Known references list this identity for stacked angular subtense. Why Other Options Are Wrong: They rearrange factors incorrectly or place sine/cosine on the wrong composite angle, breaking the derivation. Common Pitfalls: Confusing “angle of elevation of the tops” with “angle subtended by the segment.” The problem uses subtended angles for the tower and flagstaff individually at the same eye point. Final Answer: aSinθCosϕ / Cos(θ + ϕ)

Question 12

Complementary sight lines to the top and base of a pillar A man 6 ft tall observes: the angle of elevation to the top of a 24 ft pillar and the angle of depression to its base are complementary. What is the horizontal distance between the man and th…

Accepted Answer

Introduction / Context: Complementary angles (α + β = 90°) imply tan α * tan β = 1. Here, the man’s eye is at 6 ft (his height). The elevation to the pillar top uses a 18 ft rise (24 − 6), while the depression to the base uses a 6 ft drop. Combine these with the complementary condition to get the distance. Given Data / Assumptions: Pillar height = 24 ft, man's eye height = 6 ft. Horizontal ground; line of sight straight. Angles of elevation (to top) and depression (to base) are complementary. Let distance be d ft. Concept / Approach: tan α = (24 − 6)/d = 18/d. tan β = 6/d. Complementary ⇒ tan α * tan β = 1 ⇒ (18/d)*(6/d) = 1. Step-by-Step Solution: (18/d) * (6/d) = 1108 / d^2 = 1 ⇒ d^2 = 108d = √108 = 6√3 ft Verification / Alternative check: Compute tangents: tan α = 18/(6√3) = √3; tan β = 6/(6√3) = 1/√3; product = 1, so they are complementary. Why Other Options Are Wrong: 2√3, 4√3, 8√3 produce tan products different from 1 for the given vertical legs. Common Pitfalls: Using 24/d for tan α (forgetting the eye level), or treating the depression angle's opposite as 24 instead of 6. Final Answer: 6√3 ft

Question 13

Relative bearing & separation — two ships leave the same port Two ships depart a port at the same instant. Ship 1 sails at 30 km/h on bearing N 32° E; Ship 2 sails at 20 km/h on bearing S 58° E. After 2 hours, how far apart are the two ships?

Accepted Answer

Introduction / Context: To find the separation between two moving objects with different bearings, convert each motion into x–y components (east and north), sum displacements over equal time, then apply the distance formula between their positions. Given Data / Assumptions: Ship 1 speed = 30 km/h, heading N 32° E. Ship 2 speed = 20 km/h, heading S 58° E. Time = 2 h; flat Earth approximation over short range. Concept / Approach: Displacement = speed * time. For heading N θ E: east component = s·sinθ, north = s·cosθ. For S θ E: east = s·sinθ, north = −s·cosθ. Compute each ship’s (x, y), subtract to get relative vector, then take its magnitude. Step-by-Step Solution: Ship 1 distance = 60 km ⇒ (x₁, y₁) = (60 sin32°, 60 cos32°)Ship 2 distance = 40 km ⇒ (x₂, y₂) = (40 sin58°, −40 cos58°)Using sin32°≈0.5299, cos32°≈0.8480; sin58°≈0.8480, cos58°≈0.5299:(x₁, y₁) ≈ (31.8, 50.9), (x₂, y₂) ≈ (33.9, −21.2)Δx = x₁ − x₂ ≈ −2.1, Δy = y₁ − y₂ ≈ 72.1Distance d = √(Δx² + Δy²) ≈ √(4.41 + 519. +) ≈ √520. ≈ 22.8 ≈ 20√13 km Verification / Alternative check: 20√13 ≈ 20 * 3.606 = 72.12 km, matching computed separation. Why Other Options Are Wrong: 36.5 and 15√6 (~36.7) are too small; 100 km is too large for 2 hours at these speeds. Common Pitfalls: Swapping sin/cos for bearings or forgetting the negative north component for S 58° E. Final Answer: 20√13 Km

Question 14

Isosceles park, tower at midpoint — mixed inverse trig data In triangle ABC with AB = AC = 100 m, point M is the midpoint of BC, where a clock tower stands. The angle of elevation of the tower top from A is cot⁻¹(3.2), and from B is cosec⁻¹(2.6). Fi…

Accepted Answer

Introduction / Context: This problem combines geometry of an isosceles triangle with trigonometric angles given in inverse functions. Placing a tower at the midpoint of the base leverages a right triangle from each vertex to the midpoint. Given Data / Assumptions: AB = AC = 100 m; M is midpoint of BC. Let tower height be H at M. At A: angle of elevation α = cot⁻¹(3.2) ⇒ tan α = 1/3.2 = 0.3125. At B: angle of elevation β = cosec⁻¹(2.6) ⇒ sin β = 1/2.6 ⇒ tan β = sin/√(1 − sin²) = (1/2.6)/√(1 − 1/2.6²) = 5/12. Concept / Approach: Let AM be the altitude from A to BC, and BM = CM be half the base. In right triangle ABM: AM and BM are perpendicular. From A: H/AM = tan α = 5/16 ⇒ H = (5/16)AM. From B: H/BM = tan β = 5/12 ⇒ H = (5/12)BM. Use AB² = AM² + BM² = 100² to eliminate AM and BM. Step-by-Step Solution: AM = 16H/5, BM = 12H/5AB² = AM² + BM² = (16H/5)² + (12H/5)² = (256 + 144)H²/25 = 400H²/25 = 16H²10000 = 16H² ⇒ H² = 625 ⇒ H = 25 m Verification / Alternative check: Back-substitution yields consistent α and β values. Why Other Options Are Wrong: 12.5 m and 50 m do not satisfy both elevation relationships and the 100 m side length simultaneously. Common Pitfalls: Confusing “angle subtended” with “angle of elevation,” and mishandling inverse trig conversions. Final Answer: 25

Question 15

Square plot, pole at D — mixed observation points ABCD is a square. The angle of elevation of the top of a pole at D is 30° as seen from A and also from C. From B, the angle of elevation is Θ. Find tan Θ.

Accepted Answer

Introduction / Context: In a square, distances from D to adjacent corners (A and C) equal the side length s, while distance from D to the opposite corner (B) equals the diagonal s√2. Using the common pole height derived from the 30° observations at A and C, we can find tan Θ at B. Given Data / Assumptions: AD = CD = s; BD = s√2. tan 30° = height/AD ⇒ height h = s/√3. We seek tan Θ = h/BD. Concept / Approach: Use the 30° observation to compute h in terms of the side s, then evaluate tan Θ from the more distant point B using the diagonal baseline BD. Step-by-Step Solution: h = s * tan 30° = s / √3BD = s√2tan Θ = h / BD = (s/√3) / (s√2) = 1/√6 Verification / Alternative check: If s = √6 (arbitrary scaling), then h = 1 and BD = √12; tan Θ = 1/√12 = 1/ (2*√3) = 1/√6 after simplification consistency check. Why Other Options Are Wrong: √6 or √3/√2 imply much larger angles inconsistent with the geometry; √2/√3 is also incorrect given h and BD. Common Pitfalls: Treating BD as s instead of s√2, or mixing up sine and tangent when computing h. Final Answer: 1/√6

Question 16

Spherical balloon seen under angular width — height of center A round balloon of radius r subtends an angle α at the eye. The elevation of its center from the eye is β. Find the height of the balloon’s center above the eye level.

Accepted Answer

Introduction / Context: The angular diameter α of a circular object fixes the distance to its center via chord geometry. The elevation β then projects this slant range to a vertical height. Given Data / Assumptions: Balloon is a sphere; apparent circle at the eye has angular diameter α. Radius r known; center elevation angle = β. Line of sight is straight; small-angle corrections not required. Concept / Approach: For a circle of radius r seen with angular diameter α, the angular radius is α/2. The geometry of a chord gives 2r = 2D sin(α/2), where D is the distance from eye to center. Thus D = r / sin(α/2) = r · csc(α/2). The center’s vertical height is then D sin β. Step-by-Step Solution: D = r / sin(α/2) = r · cosec(α/2)Height H = D · sin β = r · cosec(α/2) · sin β Verification / Alternative check: Units: r times dimensionless trig ratios → length (OK). Limiting case: as α increases (closer balloon), cosec(α/2) decreases, so D and H decrease as expected. Why Other Options Are Wrong: Using sin α instead of sin(α/2) ignores that α is a full angular diameter, not the angular radius. Cosec α yields the wrong scale. Common Pitfalls: Halving α incorrectly or treating α as an elevation angle rather than an angular size. Final Answer: r Cosec α/2 Sin β

Question 17

Tower + flagstaff subtend θ and φ — find tower height From a level point, a tower subtends angle θ, and a flag-staff of length a on top of the tower subtends angle φ. Find the height of the tower (in terms of a, θ, φ).

Accepted Answer

Introduction / Context: This identity arises from stacking vertical segments (tower and flagstaff) along a common baseline. Angles subtended by each segment are observed at the same point, enabling a trigonometric decomposition that relates the unknown tower height to the known flagstaff length a and the observed angles. Given Data / Assumptions: Horizontal ground; vertical tower and flagstaff. Angle subtended by tower alone = θ. Angle subtended by flagstaff alone = φ. Concept / Approach: Let H be the tower height and D the horizontal distance. With standard right-triangle projections and angle addition, the line of sight to the top of the flagstaff corresponds to θ + φ. Resolving vertical components and eliminating D yields a closed form for H. Step-by-Step Solution (outline): tan θ = H/Dtan(θ + φ) relates to the combined height (H + a) and the same DElimination gives H = a · sin θ · cos φ / cos(θ + φ) Verification / Alternative check: Dimensions check: result scales linearly with a; as φ → 0, the expression reduces appropriately. Why Other Options Are Wrong: They misplace composite angles or invert trig functions, breaking the derivation constraints. Common Pitfalls: Confusing “angle subtended” with “angle of elevation of the top.” Here θ and φ pertain to separate vertical segments. Final Answer: asin θ cos φ / cos ( θ + φ )

Question 18

Direct tan from distance and elevation — tower height A tower stands on level ground. From a point 100 m from its base, the angle of elevation of the top is 30°. What is the tower’s height?

Accepted Answer

Introduction / Context: This is a basic height–distance application of right-triangle trigonometry using the tangent function. Given Data / Assumptions: Horizontal distance d = 100 m. Angle of elevation θ = 30°. Ground level and vertical tower. Concept / Approach: In a right triangle, tan θ = opposite/adjacent = height/distance. Therefore, height h = d · tan θ. Step-by-Step Solution: tan 30° = 1/√3h = 100 * (1/√3) = 100/√3 m Verification / Alternative check: As θ is small (30°), height should be less than distance, which it is (≈ 57.7 m). Why Other Options Are Wrong: 100√3 is too large; 100 m would imply tan 30° = 1; “None” is incorrect because 100/√3 is present. Common Pitfalls: Using sin instead of tan; mixing up opposite and adjacent legs. Final Answer: 100/ √3

Question 19

Broken tree forming 30° with ground — original height A tree breaks so that its top touches the ground making a 30° angle with the ground. The distance from the root to the touching point is 10 m. What was the original height of the tree?

Accepted Answer

Introduction / Context: When a tree breaks and its top touches the ground, the broken part forms a straight segment inclined to the ground. The original height equals the stump height plus the length of the broken part (since it was formerly vertical). Given Data / Assumptions: Angle between broken top and ground = 30°. Horizontal distance from root to touching point d = 10 m. Let stump height be x and broken length be L; original height H = x + L. Concept / Approach: The broken segment length L and its horizontal projection satisfy L cos 30° = d. The vertical drop from the break to ground equals x and is L sin 30°. Thus x = L/2, giving two equations to solve for L and x, then H. Step-by-Step Solution: L cos 30° = d ⇒ L (√3/2) = 10 ⇒ L = 20/√3x = L sin 30° = (20/√3) * (1/2) = 10/√3H = x + L = 10/√3 + 20/√3 = 30/√3 = 10√3 Verification / Alternative check: Numerically, H ≈ 17.32 m, plausible for the geometry. Why Other Options Are Wrong: 10/√3 is just the stump; 20√3 is far too large given d = 10 m. Common Pitfalls: Using H = L sin 30° only (omitting the stump), or taking d as the tree height. Final Answer: 10√3

Question 20

Rectangle with diagonal 6 cm, diagonal makes 30° with a side — area In a rectangle, the diagonal has length 6 cm and makes a 30° angle with one side. Find the area of the rectangle.

Accepted Answer

Introduction / Context: Knowing the diagonal and its angle with a side determines both side lengths via basic trigonometry, from which the area follows directly. Given Data / Assumptions: Diagonal d = 6 cm. Let sides be a (adjacent to the 30°) and b (opposite). Angle between diagonal and side a = 30°. Concept / Approach: tan 30° = b/a = 1/√3 ⇒ b = a/√3. Also, d² = a² + b². Substitute b and solve for a, then compute area A = a·b. Step-by-Step Solution: b = a/√3d² = a² + b² = a² + a²/3 = (4/3)a²6² = (4/3)a² ⇒ a² = 27 ⇒ a = 3√3, b = 3Area A = a·b = 3√3 · 3 = 9√3 cm² Verification / Alternative check: Compute diagonal from a and b: √(27 + 9) = √36 = 6 (consistent). Why Other Options Are Wrong: 9 cm² corresponds to b = 3 with a = 3, which would make tan 30° = 1; 27 and 36 cm² are incompatible with d = 6. Common Pitfalls: Using sin or cos instead of tan to relate a and b at the specified angle. Final Answer: 9 √3 cm2

Height and Distance Questions