Question

A horizontal distance between two poles is 15 m the angle of depression of the top of the first pole as seen from the top of the second is 30 degree if the height of the second pole is 24 m find the height of the first pole

Answer 1

Answer:

Step-by-step explanation:

Height of the first pole is

3(8-5\/3) m

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.