Question

The horizontal distance between two poles is 15m. The angles of depression of the top of the first pole as seen from the top of the second pole is 30°. If the height of the second pole is 24m,find the height of the first pole.

Answer 1

SOLUTION

Let the height of the first pole be AB=h m &

of second pole be CE (CD+DE) =24m.

(let AB||DE) & AD||DE)

& AD=BE= 15m

AB= DE=h m

BE=AD= 15m

CE= 24m & CD=24-h m

In ∆CDA,

$tan30 \degree = \frac{24 - h}{15} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{24 - h}{15} \\ \\ = > \frac{15}{ \sqrt{3} } = 24 - h \\ \\ = > \frac{15}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } = > \frac{15 \sqrt{3} }{3} \\ \\ = > 5 \sqrt{3} = 24 - h \\ \\ = > h = 24 - 5 \sqrt{3} \\ \\ = > h = 24 - 5 \times 1.732 \: \: \: \: \: \: ( \sqrt{3} = 1.732) \\ \\ = > h = 24 - 8.66 \\ \\ = > h = 15.34m$

Hope it helps ☺️