Question

There is pillar QP standing on a tower PN. There is a point A at a horizontal distance of 40
m from the base N of the tower. Tower PN and pillar QP substand angle $\theta$ and (phi)
respectively at the point A such that tan [tex]\theta = 1/2 and tan phi = .2/3 Find the height of the pillar.


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Answer 1

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$\rule{300}2$

◇ Solution ◇

let the height of the tower be 'H' metre and height of the pillar ne 'h' metre

in ∆ APN,

$tan\theta= \frac{PN}{AN} \\ or \frac{1}{2}= \frac{H}{40} \\ H=\frac{40}{2}=20$

in ∆AQN.

$tan\theta = \frac{QN}{AN} \\ \frac{2}{3}= \frac{h+H}{40} \\ 3×(h+20)=40×2 \\ 3h=80-60=20 \\ h=\frac{20}{3}=6.67$

$\rule{300}2$