Let AB and PQ be two vertical poles, 160 m apart from each other.

Question

Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let $${\pi  \over 8}$$ and $$\theta$$ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then tan2$$\theta$$ is equal to

Accepted Answer

$${l \over {80}} = \tan \theta $$ ..... (i)

$${{2l} \over {80}} = \tan {\pi  \over 8}$$ ..... (ii)

From (i) and (ii)

$${1 \over 2} = {{\tan \theta } \over {\tan {\pi  \over 8}}} \Rightarrow {\tan ^2}\theta  = {1 \over 4}{\tan ^2}{\pi  \over 8}$$

$$ \Rightarrow {\tan ^2}\theta  = {{\sqrt 2  - 1} \over {4(\sqrt 2  + 1)}} = {{3 - 2\sqrt 2 } \over 4}$$

Solutions · 1