Height and Distance: Solved Examples

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

→ 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?

→ 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.

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

→ √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

→ 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

→ 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)

→ 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

→ √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