Question

The angle of elevation of the top B of a tower AB from a point X on the ground is 60. At a point Y, 40m vertically above X, the angle of elevation is 45. Find the height of the tower AB and the distance XB.

Answer 1

hope i helped u......:)

Answer 2

Answer:

94.641 m

Step-by-step explanation:

Refer the attached figure

AB is the height of tower

XY = 40 m = BC

The angle of elevation of the top B of a tower AB from a point X on the ground is 60.i.e.∠BXA = 60°

At a point Y, 40m vertically above X, the angle of elevation is 45.i.e.∠BYC = 45°

Let BC be x

AB = 40+x

In ΔBYC

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

$tan 45^{\circ}= \frac{BC}{YC}$

$1= \frac{x}{YC}$

$YC =x$

So, YC = XA = x

In ΔAXB

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

$tan 60^{\circ}= \frac{AB}{XA}$

$\sqrt{3}= \frac{x+40}{x}$

$\sqrt{3}x= x+40$

$\sqrt{3}x-x=40$

$(\sqrt{3}-1)x=40$

$x=\frac{40}{(\sqrt{3}-1)}$

$x=54.641$

So, AB = 40+54.641 =94.641

Hence the height of the tower is 94.641 m