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An observer of height 1.8 m is 13.2 m away from a palm tree. The angle of elevation of the top of the tree from his eyes is $45^{\circ}$ . What is the height of the palm tree?

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answer is C.

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Detailed Solution

Given a palm tree observer of height of

AE

= 1.8 m

ED = AC = 13.2 m

Now, In the triangle

△ ABC

\tan 45^{\circ} = \frac{BC}{AC}

1 = \frac{BC}{13.2}

BC = 13.2 m

The height of the palm tree = BD.
Then,

BD = BC + CD

BD = 13.2 + AE

BD = 13.2 + 1.8

BD = 15 m

Hence the height of the palm tree is equal to 15 m.
Option 3 is correct.

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