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

Answer 1

Answer:

Step-by-step explanation:

the triangle ABP s isosceles wid centrl angle=120...AB=60..AP=BP=60/rt 3=20 rt 3...

height f tower=20 rt3/rt 3=20..!!!

Answer 2

Answer:

25.98 m

Step-by-step explanation:

Refer the attached figure

Let AQ ber x

So, BQ = 60-x

In ΔPAQ

$Tan\theta =\frac{Perpendicular}{Base}$

$Tan 30^{\circ} =\frac{PQ}{AQ}$

$\frac{1}{\sqrt{3}}=\frac{PQ}{x}$

$\frac{1}{\sqrt{3}}x=PQ$

In ΔPBQ

$Tan\theta =\frac{Perpendicular}{Base}$

$Tan 60^{\circ} =\frac{PQ}{BQ}$

$\sqrt{3}=\frac{PQ}{60-x}$

$\sqrt{3}(60-x)=PQ$

So,$\sqrt{3}(60-x)=\frac{1}{\sqrt{3}}x$

$x=45$

$\frac{1}{\sqrt{3}}(45)=PQ$

PQ = 25.98 m

Hence the height of tower is 25.98 m