Question

Two posts are k meter apart and the height of one is double that of the other. If from the mid point of the line segment joining their feet, an observer finds the angles of elevation of their tops to be complementary then find the height of the shorted post.

Answer 1

HELLO DEAR,




Let AB = k m be the line joining the feet of the persons and C be the mid-point of this line.

Thus, AC = CB = k/ 2 m

Also, let h be the height of the of the shorter person then the height of the taller person will be 2h.

$\tan(x) = \frac{h}{ \frac{k}{2} } = \frac{2h}{k} ..............(1) \\ = > \tan(90 - x) = \frac{2h}{ \frac{k}{2} } = \frac{4h}{k} \\ = > \cot(x) = \frac{4h}{k} ...............(2) \\ = > from(1)and(2)multiplying \\ = > \tan(x) \cot(x) = \frac{2h}{k} \times \frac{4h}{k} \\ = > \tan(x) \times \frac{1}{ \tan(x) } = \frac{8 {h}^{2} }{ {k}^{2} } \\ = > 8 {h}^{2} = {k}^{2} \\ = > {h}^{2} = \frac{ {k}^{2} }{ {(2 \sqrt{2}) }^{2} } \\ = > h = \frac{k}{2 \sqrt{2} } meter$

I HOPE ITS HELP YOU DEAR,
THANKS


 

Answer 2

In ABC,
tan theta = AB/BC = 2x/k...........(1)

In CDE,
cot (90-theta) = CD/DE = k/4x
tan theta = k/4x..........(2)

Equating (1) and (2), we get,
2x/k = k/4x
8x^2 = k^2
x = k / 2√2

Hope this helps....
Plsss make it the brainiest