Question

The horizons distance between two poles is 15. The angle of depression of the top of the first pole as seen from thevtopbof the second pole is 30°. If the height

Answer 1

please mention the height of either of the pole because the question is incomplete without that....

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.