Question

38. 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, 40
m vertically above X, the angle of elevation of the top is 45°. Find the height of the tower AB and the
distance XB.
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Answer 1

Answer:

hope this may help u...

first see above img

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}tanθ=BasePerpendicular

tan 45^{\circ}= \frac{BC}{YC}tan45∘=YCBC

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

YC =xYC=x

So, YC = XA = x

In ΔAXB

tan\theta= \frac{Perpendicular}{Base}tanθ=BasePerpendicular

tan 60^{\circ}= \frac{AB}{XA}tan60∘=XAAB

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

\sqrt{3}x= x+403x=x+40

\sqrt{3}x-x=403x−x=40

(\sqrt{3}-1)x=40(3−1)x=40

x=\frac{40}{(\sqrt{3}-1)}x=(3−1)40

x=54.641x=54.641

So, AB = 40+54.641 =94.641

Hence the height of the tower is 94.641 m

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