Question

The horizontal distance between two poles are 15m.The angle of depression of the top of the first pole from the top of the second pole is 30º.If the height of the second tower is 24m, find the height of the first tower.

Answer 1

Given distance between two poles is BD = 15 m
Hence CE = 15m
Given height of second pole, AB = 24 cm

Let height of first pole be CD = h
Hence BE = h
Therefore, AE = 24 – h
In right angled triangle AEC,
tan30° = (AE/CE)
(1/√3) = (24 – h)/15
15 = 24√3 - √3h
√3h = 24√3 – 15
h = 24 – (15√3/3) = 24 – 5√3
Therefore, h = 15.34 m (nearly) 

Answer 2

Let AB be the pole of height h metres and CD be the second pole of height 24m.

Draw CF||BD and AE ||BD.Then,

BD = AE = 15., CE = (24 - h) m

$\sf\angle{{CAE}=\angle{ACF}=30°}$

From right ∆AEC, we have

$\longrightarrow$$\sf{tan\:30°= \frac{CE}{AE}=\frac{1}{ \sqrt{3} } = \frac{24 - h}{15}}$

$\longrightarrow$$\sf{\frac{15}{ \sqrt{3} } = 24 - h}$

$\longrightarrow$$\sf{h = 24 - \frac{15}{ \sqrt{3} } = 24 - 5 \sqrt{3}}$

$\longrightarrow$$\sf{24-5×1.732}$

$\longrightarrow$$\sf{24-8.66}$

$\longrightarrow$${\boxed{\sf{\red{15.34m}}}}$

Hence, the height of the first pole = 15.34m.