Question

A vertical tower PQ subtends equal angle of 30° at each of the two places A and B, 60 meter apart on the ground. If AB subtends an angle of 60° at P(the foot of the tower) then the height of the tower is
(A) 20$\sqrt{3}$ meter
(B) 20 meter
(C) 60 $\sqrt{3}$ meter
(D) 60 meter
Kindly answer with explanation.

Answer 1

The height of tower=$20\sqrt 3$m

Step-by-step explanation:

Let h be the height of tower.

PQ=h

In triangle PQA

$tan\theta=\frac{Perpendicular\;side}{base}$

$tan 30=\frac{QP}{AP}$

$\frac{1}{\sqrt 3}=\frac{h}{AP}$

$AP=h\sqrt 3$

AP=BP=$h\sqrt 3$

Cosine law: $c^2=a^2+b^2-2abcos\theta$

$\theta=60^{\circ},a=AP,b=BP,c=60 m$

Substitute the values then we get

$(60)^2=(AP)^2+(BP)^2-2AP\cdot BPcos 60^{\circ}$

$3600=(h\sqrt 3)^2+(h\sqrt 3)^2-2(h\sqrt3)(h\sqrt3 )\times \frac{1}{2}$

$3600=3h^2+3h^2-6h^2\times \frac{1}{2}$

$3600=6h^2-3h^2$

$3h^2=3600$

$h^2=1200$

$h=\sqrt{1200}=20\sqrt 3$

Hence, the height of tower=$20\sqrt 3$m

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