Board Previous Year Questions (PYQs)

These are exam-style height-and-distance questions modelled on CBSE and State Board papers from recent years. Each is fully solved with a clear diagram-driven method.

Scoring tip: Draw the labelled triangle, write the tan⁡\tan equation, and keep surds exact. Show the angle transfer for any depression.

Work through all 26. They span 2-mark, 3-mark and 5-mark patterns.

PYQ 1 (2 marks): The elevation of a tower's top from a point 20 m away is 60°. Find the height.

Solution:

  1. h=20tan⁡60°=203h = 20\tan 60° = 20\sqrt3.

Final Answer: 203≈34.6420\sqrt3 \approx 34.64 m.

PYQ 2 (2 marks): A kite is at height 75 m with string at 60° to the ground. Find the string length.

Solution:

  1. sin⁡60°=75ℓ⇒ℓ=753/2=1503=503\sin 60° = \dfrac{75}{\ell} \Rightarrow \ell = \dfrac{75}{\sqrt3/2} = \dfrac{150}{\sqrt3} = 50\sqrt3.

Final Answer: 503≈86.650\sqrt3 \approx 86.6 m.

PYQ 3 (2 marks): A pole 6 m high casts a shadow 232\sqrt3 m long. Find the sun's elevation.

Solution:

  1. tan⁡θ=623=33=3⇒θ=60°\tan\theta = \dfrac{6}{2\sqrt3} = \dfrac{3}{\sqrt3} = \sqrt3 \Rightarrow \theta = 60°.

Final Answer: 60°.

PYQ 4 (2 marks): From a 100 m cliff, the depression of a boat is 45°. Find the boat's distance from the foot.

Solution:

  1. tan⁡45°=100d=1⇒d=100\tan 45° = \dfrac{100}{d} = 1 \Rightarrow d = 100 m.

Final Answer: 100 m.

PYQ 5 (3 marks): The elevations of a tower's top from two points 30 m apart in line are 30° and 60°. Find the height.

Solution:

  1. Near x=h3x = \dfrac{h}{\sqrt3}; far x+30=h3x + 30 = h\sqrt3.
  2. h3+30=h3⇒30=2h3⇒h=153\dfrac{h}{\sqrt3} + 30 = h\sqrt3 \Rightarrow 30 = \dfrac{2h}{\sqrt3} \Rightarrow h = 15\sqrt3.

Final Answer: 153≈25.9815\sqrt3 \approx 25.98 m.

PYQ 6 (3 marks): A 1.5 m tall observer is 28.5 m from a tower; the elevation of the top is 45°. Find the tower's height.

Solution:

  1. Triangle height =28.5tan⁡45°=28.5= 28.5\tan 45° = 28.5.
  2. Total =28.5+1.5=30= 28.5 + 1.5 = 30 m.

Final Answer: 30 m.

PYQ 7 (3 marks): From the top of a 7 m building, the elevation of a tower's top is 60° and the depression of its foot is 45°. Find the tower's height.

Solution:

  1. Depression 45°: d=7d = 7.
  2. Elevation 60°: rise =7tan⁡60°=73= 7\tan 60° = 7\sqrt3.
  3. Height =7+73=7(1+3)= 7 + 7\sqrt3 = 7(1 + \sqrt3).

Final Answer: 7(1+3)≈19.127(1 + \sqrt3) \approx 19.12 m.

PYQ 8 (3 marks): From a 75 m lighthouse, the depressions of two ships in line are 30° and 45°. Find the distance between them.

Solution:

  1. Near x1=75x_1 = 75; far x2=753x_2 = 75\sqrt3.
  2. Distance =753−75=75(3−1)= 75\sqrt3 - 75 = 75(\sqrt3 - 1).

Final Answer: 75(3−1)≈54.975(\sqrt3 - 1) \approx 54.9 m.

PYQ 9 (3 marks): The elevation of a tower from a point is 30°; on walking 60 m towards it the elevation is 60°. Find the tower's height.

Solution:

  1. Near x=h3x = \dfrac{h}{\sqrt3}; far x+60=h3x + 60 = h\sqrt3.
  2. h3+60=h3⇒60=2h3⇒h=303\dfrac{h}{\sqrt3} + 60 = h\sqrt3 \Rightarrow 60 = \dfrac{2h}{\sqrt3} \Rightarrow h = 30\sqrt3.

Final Answer: 303≈51.9630\sqrt3 \approx 51.96 m.

PYQ 10 (3 marks): A ladder 15 m long just reaches a window 9 m above the ground. Find the distance of the foot of the ladder from the wall.

Solution:

  1. Foot distance =152−92=225−81=144=12= \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12 m.

Final Answer: 12 m.

PYQ 11 (3 marks): The shadow of a tower is 30 m longer when the sun's elevation is 30° than at 60°. Find the tower's height.

Solution:

  1. h3−h3=30⇒2h3=30⇒h=153h\sqrt3 - \dfrac{h}{\sqrt3} = 30 \Rightarrow \dfrac{2h}{\sqrt3} = 30 \Rightarrow h = 15\sqrt3.

Final Answer: 153≈25.9815\sqrt3 \approx 25.98 m.

PYQ 12 (3 marks): An aeroplane at height 1200 m observes two ships in line at depressions 60° and 30°. Find the distance between the ships.

Solution:

  1. Near x1=1200tan⁡60°=12003=4003x_1 = \dfrac{1200}{\tan 60°} = \dfrac{1200}{\sqrt3} = 400\sqrt3.
  2. Far x2=1200tan⁡30°=12003x_2 = \dfrac{1200}{\tan 30°} = 1200\sqrt3.
  3. Distance =12003−4003=8003= 1200\sqrt3 - 400\sqrt3 = 800\sqrt3.

Final Answer: 8003≈1385.6800\sqrt3 \approx 1385.6 m.

PYQ 13 (3 marks): Two poles of equal height stand on either side of a 80 m wide road. From a point between them the elevations of their tops are 60° and 30°. Find the height and the point's position.

Solution:

  1. tan⁡60°=hx\tan 60° = \dfrac{h}{x}, tan⁡30°=h80−x\tan 30° = \dfrac{h}{80 - x}.
  2. h=x3=80−x3⇒3x=80−x⇒x=20h = x\sqrt3 = \dfrac{80 - x}{\sqrt3} \Rightarrow 3x = 80 - x \Rightarrow x = 20.
  3. h=203h = 20\sqrt3.

Final Answer: Height 203≈34.6420\sqrt3 \approx 34.64 m; point 20 m from the 60° pole.

PYQ 14 (3 marks): A tree breaks and the top touches the ground, making 30° with the ground at a distance of 838\sqrt3 m from the foot. Find the original height of the tree.

Solution:

  1. Broken (slant) part: cos⁡30°=83ℓ⇒ℓ=833/2=16\cos 30° = \dfrac{8\sqrt3}{\ell} \Rightarrow \ell = \dfrac{8\sqrt3}{\sqrt3/2} = 16.
  2. Standing part: tan⁡30°=p83⇒p=83⋅13=8\tan 30° = \dfrac{p}{8\sqrt3} \Rightarrow p = 8\sqrt3\cdot\dfrac{1}{\sqrt3} = 8.
  3. Original height =p+ℓ=8+16=24= p + \ell = 8 + 16 = 24 m.

Final Answer: 24 m.

PYQ 15 (5 marks): From the top of a 50 m building, the elevation of the top of a tower is 60° and the depression of its foot is 30°. Find the height of the tower and the distance between the building and the tower.

Solution:

  1. Depression 30°: tan⁡30°=50d⇒d=503\tan 30° = \dfrac{50}{d} \Rightarrow d = 50\sqrt3.
  2. Elevation 60°: rise b=dtan⁡60°=503⋅3=150b = d\tan 60° = 50\sqrt3\cdot\sqrt3 = 150.
  3. Tower height =50+150=200= 50 + 150 = 200 m.

Final Answer: Tower height 200 m; distance 503≈86.650\sqrt3 \approx 86.6 m.

PYQ 16 (5 marks): The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud above the lake.

Solution:

  1. Let cloud height above lake =H= H, horizontal distance =d= d. Observation point is 60 m above the lake.
  2. Elevation: tan⁡30°=H−60d⇒d=(H−60)3\tan 30° = \dfrac{H - 60}{d} \Rightarrow d = (H - 60)\sqrt3.
  3. Reflection is HH below the lake, so depression: tan⁡60°=H+60d⇒d=H+603\tan 60° = \dfrac{H + 60}{d} \Rightarrow d = \dfrac{H + 60}{\sqrt3}.
  4. (H−60)3=H+603⇒3(H−60)=H+60⇒2H=240⇒H=120(H - 60)\sqrt3 = \dfrac{H + 60}{\sqrt3} \Rightarrow 3(H - 60) = H + 60 \Rightarrow 2H = 240 \Rightarrow H = 120.

Final Answer: 120 m above the lake.

PYQ 17 (5 marks): A man on a 100 m cliff observes a boat's depression change from 30° to 60° in 2 minutes. Find the speed of the boat.

Solution:

  1. Far x2=100tan⁡30°=1003x_2 = \dfrac{100}{\tan 30°} = 100\sqrt3. Near x1=100tan⁡60°=1003=10033x_1 = \dfrac{100}{\tan 60°} = \dfrac{100}{\sqrt3} = \dfrac{100\sqrt3}{3}.
  2. Distance =1003−10033=20033= 100\sqrt3 - \dfrac{100\sqrt3}{3} = \dfrac{200\sqrt3}{3} m.
  3. Speed =2003/32=10033≈57.7= \dfrac{200\sqrt3/3}{2} = \dfrac{100\sqrt3}{3} \approx 57.7 m/min.

Final Answer: 10033≈57.7\dfrac{100\sqrt3}{3} \approx 57.7 m/min.

PYQ 18 (5 marks): The elevation of the top of a tower from two points at distances aa and bb from the base (in line, on the same side) are complementary. Prove the height is ab\sqrt{ab}.

Solution:

  1. Let the angles be θ\theta (at distance aa) and 90°−θ90° - \theta (at distance bb).
  2. tan⁡θ=ha\tan\theta = \dfrac{h}{a} and tan⁡(90°−θ)=cot⁡θ=hb\tan(90° - \theta) = \cot\theta = \dfrac{h}{b}.
  3. Multiply: tan⁡θcot⁡θ=h2ab⇒1=h2ab⇒h=ab\tan\theta\cot\theta = \dfrac{h^2}{ab} \Rightarrow 1 = \dfrac{h^2}{ab} \Rightarrow h = \sqrt{ab}.

Final Answer: h=abh = \sqrt{ab}. Proved.

PYQ 19 (5 marks): A tower stands on a 20 m building. From a point on the ground, the elevation of the bottom of the tower is 45° and of the top is 60°. Find the tower's height.

Solution:

  1. tan⁡45°=20d⇒d=20\tan 45° = \dfrac{20}{d} \Rightarrow d = 20.
  2. tan⁡60°=20+t20=3⇒20+t=203\tan 60° = \dfrac{20 + t}{20} = \sqrt3 \Rightarrow 20 + t = 20\sqrt3.
  3. t=20(3−1)t = 20(\sqrt3 - 1).

Final Answer: 20(3−1)≈14.6420(\sqrt3 - 1) \approx 14.64 m.

PYQ 20 (5 marks): The angles of depression of the top and bottom of an 8 m building from the top of a multi-storeyed building are 30° and 45°. Find the height of the multi-storeyed building and the distance between the two buildings.

Solution:

  1. Let the tall building height =H= H, distance =d= d.
  2. Bottom (45°): tan⁡45°=Hd⇒d=H\tan 45° = \dfrac{H}{d} \Rightarrow d = H.
  3. Top (30°): the drop to the 8 m building's top is H−8H - 8, so tan⁡30°=H−8d=H−8H\tan 30° = \dfrac{H - 8}{d} = \dfrac{H - 8}{H}.
  4. 13=H−8H⇒H=3(H−8)⇒H(3−1)=83⇒H=833−1=43(3+1)=12+43\dfrac{1}{\sqrt3} = \dfrac{H - 8}{H} \Rightarrow H = \sqrt3(H - 8) \Rightarrow H(\sqrt3 - 1) = 8\sqrt3 \Rightarrow H = \dfrac{8\sqrt3}{\sqrt3 - 1} = 4\sqrt3(\sqrt3 + 1) = 12 + 4\sqrt3.

Final Answer: H=12+43≈18.93H = 12 + 4\sqrt3 \approx 18.93 m; distance =H=12+43= H = 12 + 4\sqrt3 m.

PYQ 21 (3 marks): The elevation of the top of a hill from the foot of a tower is 60°, and the elevation of the top of the tower (height 50 m) from the foot of the hill is 30°. Find the height of the hill.

Solution:

  1. From the hill foot, tower top: tan⁡30°=50d⇒d=503\tan 30° = \dfrac{50}{d} \Rightarrow d = 50\sqrt3.
  2. From the tower foot, hill top: tan⁡60°=Hd⇒H=503⋅3=150\tan 60° = \dfrac{H}{d} \Rightarrow H = 50\sqrt3\cdot\sqrt3 = 150.

Final Answer: Hill height =150= 150 m.

PYQ 22 (3 marks): From the top of a 120 m tower, the depressions of two cars on the same side are 60° and 45°. Find the distance between the cars.

Solution:

  1. Near x1=120tan⁡60°=1203=403x_1 = \dfrac{120}{\tan 60°} = \dfrac{120}{\sqrt3} = 40\sqrt3.
  2. Far x2=120tan⁡45°=120x_2 = \dfrac{120}{\tan 45°} = 120.
  3. Distance =120−403=40(3−3)= 120 - 40\sqrt3 = 40(3 - \sqrt3).

Final Answer: 40(3−3)≈50.740(3 - \sqrt3) \approx 50.7 m.

PYQ 23 (3 marks): A statue 1.6 m tall stands on a pedestal. From a point on the ground, the elevation of the top of the statue is 60° and of the top of the pedestal is 45°. Find the pedestal's height.

Solution:

  1. Let pedestal height pp, distance dd. tan⁡45°=pd⇒d=p\tan 45° = \dfrac{p}{d} \Rightarrow d = p.
  2. tan⁡60°=p+1.6d=p+1.6p=3\tan 60° = \dfrac{p + 1.6}{d} = \dfrac{p + 1.6}{p} = \sqrt3.
  3. p+1.6=p3⇒1.6=p(3−1)⇒p=1.63−1=0.8(3+1)p + 1.6 = p\sqrt3 \Rightarrow 1.6 = p(\sqrt3 - 1) \Rightarrow p = \dfrac{1.6}{\sqrt3 - 1} = 0.8(\sqrt3 + 1).

Final Answer: p=0.8(3+1)≈2.18p = 0.8(\sqrt3 + 1) \approx 2.18 m.

PYQ 24 (5 marks): From a point on the ground, the elevation of an aeroplane is 60°. After a flight of 30 seconds at a constant height of 300033000\sqrt3 m, the elevation becomes 30°. Find the speed of the aeroplane.

Solution:

  1. Near x1=30003tan⁡60°=300033=3000x_1 = \dfrac{3000\sqrt3}{\tan 60°} = \dfrac{3000\sqrt3}{\sqrt3} = 3000.
  2. Far x2=30003tan⁡30°=30003⋅3=9000x_2 = \dfrac{3000\sqrt3}{\tan 30°} = 3000\sqrt3\cdot\sqrt3 = 9000.
  3. Distance =9000−3000=6000= 9000 - 3000 = 6000 m in 30 s.
  4. Speed =600030=200= \dfrac{6000}{30} = 200 m/s.

Final Answer: 200 m/s (= 720 km/h).

PYQ 25 (3 marks): The elevation of the top of a vertical tower from a point on the ground is 45°. From a point 10 m vertically above the first, the elevation is 30°. Find the height of the tower.

Solution:

  1. From ground: tan⁡45°=hd⇒d=h\tan 45° = \dfrac{h}{d} \Rightarrow d = h.
  2. From 10 m up: tan⁡30°=h−10d=h−10h\tan 30° = \dfrac{h - 10}{d} = \dfrac{h - 10}{h}.
  3. 13=h−10h⇒h=3(h−10)⇒h(3−1)=103⇒h=1033−1=53(3+1)=15+53\dfrac{1}{\sqrt3} = \dfrac{h - 10}{h} \Rightarrow h = \sqrt3(h - 10) \Rightarrow h(\sqrt3 - 1) = 10\sqrt3 \Rightarrow h = \dfrac{10\sqrt3}{\sqrt3 - 1} = 5\sqrt3(\sqrt3 + 1) = 15 + 5\sqrt3.

Final Answer: 15+53≈23.6615 + 5\sqrt3 \approx 23.66 m.

PYQ 26 (5 marks): Two ships are on opposite sides of a 100 m lighthouse. The angles of depression from the top are 30° and 45°. Find the distance between the two ships.

Solution:

  1. One side (30°): x1=100tan⁡30°=1003x_1 = \dfrac{100}{\tan 30°} = 100\sqrt3.
  2. Other side (45°): x2=100tan⁡45°=100x_2 = \dfrac{100}{\tan 45°} = 100.
  3. Distance =x1+x2=1003+100=100(3+1)= x_1 + x_2 = 100\sqrt3 + 100 = 100(\sqrt3 + 1).

Final Answer: 100(3+1)≈273.2100(\sqrt3 + 1) \approx 273.2 m.