Height and Distance: Solved Examples

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Example 1: The height of a tower is 50m. The angle of elevation of the top of the tower from a point on A on the ground is 300. Find the distance between the bottom of the tower and point A.
Sol: Here BC is tower of height 50m. In this question, we have to find the
Distance AB. Now in ΔABC, tan300 = BC/AB
Height And Distances Solved
→1/√3 = 50/AB
→ AB=50√3 m
Let us see one more example where the horizontal distance is given and the height of the object is asked.
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Example 2: While standing on the ground at a point A, if one looks at the top of a building the angle of elevation is 600. If A is 32m from the building, find the height of the building.
Sol: Here BC is the building and the person is standing at A such that
AB = 32m. In ΔABC, tan300 = BC/AB
Height And Distances Solved
→ √3 = BC/AB → BC=32√3 m
Hence the height of the building is 32√3m.
Next there are problems where a person observes two ships and the angles of depression/elevation are given and we have to find the distance between the ships etc.
Example 3: From the top of a light house 80m high, two ships are observed. The angles of elevation of these ships as seen from the light house are found to be 450 and 300. Find the distance between the two ships.
Sol: Here AB is the light house and the ships are at points
C and D. Let CD = x
In ΔABC, tan450 =AB/AC
Height And Distances Solved
→ 1 = 80/AC
→ AC=80m
Hence AD = 80 + x
In ΔBAD, tan300 =AB/AD
→ 1/√3 = 80/(80+x)
→ (80+x) = 80√3
→ x=80(√3-1) = 80 x 0.732 =58.56m
Hence the distance between the ships is 58.56 m.
Example 4: On the roof of a building 20 m high, a boy is flying a kite. To keep that kite flying a thread of length 75 m is released. If the thread makes an angle of 450 with the horizontal, then at what height from the ground is the kite flying?
Sol: Let AB is the building of height 20 m and BC is the string.
In ΔBDC, sin450 =CD/BC

Height And Distances Solved
→ 1/√2 = CD/BC
→ 1/√2 = CD/75
→ CD = 75/√2 = 43.3m
Hence the distance above the ground = 20 + 43.3 = 63.3 m
Example 5: Two poles of height 30 meters and 10 meters are erected on the ground. The top of the poles are tied by a wire of length 40 m. Find the angle made by the wire with the horizontal.
Sol: Let AB and CD are the poles of heights 30 m and 10 m respectively.
Let the angle made by the wire with horizontal is θ.
Height And Distances Solved
In ΔDEB, sinθ = BE/BD =20/40 = 1/2
→ sinθ = 300
Example 6: The angles of depression of the top and the bottom of a tree as seen from the top of a tower 90 meters high are 30 and 60 respectively. Find the height of the tree.
Sol: Let AB is the tower of height 90 m and CD is the tree.
In ΔABC, tan600 =AB/AC
Height And Distances Solved
→ √3 = 90/AC
→ AC = 30√3
Hence DE = AC = 30√3m
In ΔDEB, tan300 =BE/DE
→ 1/√3 = BE/30√3
→BE = 30m
Hence CD = AE = 90 – 30 = 60m.
Example 7: The upper part of a tree is broken and the top touches the ground making an angle of 30°. The distance of the point where the top touches the ground to the base of the tree is 72 m. What was the height of the tree?
Sol: Let AC is the tree which broke at point B so that the top C touches the ground. Here AC = 72 m
In ΔABC, tan300 = AB/AC
Height And Distances Solved
→ 1/√3 = AB/72
→ AB = 72/√3 = 24√3m
Again in ΔABC, sin300 = AB/AC
→ 1/2 = 24√3/BC
→ BC = 48√3m
Hence the length of the tree = AB + BC = 24√3+48√3 = 72√3 = 124.7m
Example 8: Two aero planes flying in opposite directions are at points A and B, 3000 m high from the ground. When the pilots look at a point P on the ground, the angles of depression made by them are 300 and 600 respectively. Find the distance between the two aero planes.
Sol: The planes are at point A & B.
In ΔADP, AD/DP =tan 300
Height And Distances Solved
→ 3000/DP =1/√3 → DP = 3000√3m
Again in ΔBPC, tan 600=3000/PC
→ √3 = 3000/PC → PC = 3000/√3
The distance between the planes = 3000√3 + 3000/√3 = 12000/√3 =6928.4 m
Example 9: When a man looks from the foot and the top of a tower at the roof of a building, the angles of elevation and depression are of 270 and 630 respectively. If the height of the building is 40m, find the height of the tower. (tan 630=2)
Sol: Let AB is the tower and CD is the building of height 40m.
In ΔACD, AC/DC = cot270 = cot(90-63)
Height And Distances Solved
→ AC/40 = tan630 = 2
→ AC=80m
Now DE=AC=80m
In ΔBED, tan 630= BE/DE
→2=BE/80 →BE=160m ∴Height of the tower = AE + EB = 40 + 160 = 200m
Example 10: There is a cliff 80 m high on the bank of a lake. A man is sitting on the cliff. He looks at a bird in the sky and the angle of elevation is formed to be of 300. When he looks at the shadow of the bird in the lake, the angle of depression is 600. Find the height of the bird from the lake surface.
Sol: Let AB is the cliff and the man is sitting at B. Let C is the position of the bird and D is its shadow.
In ΔADB, tan 600 = AB/AD
Height And Distances Solved
→ √3 = 80/AD → AD=80/√3 m
Now BE = AD = 80/√3 m
In ΔBEC, tan 300= CE/BE
→ 1/√3 = CE/(80/√3) → CE=80/3 m
∴ The height of the bird = DC = DE +CE = 80 + 80/3 m = 320/3 m
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