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JAMB Mathematics 2024 Paper

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Question : 17

Total: 50

From the top of a building 10m high, the angle of elevation of a fruit on top of a tree 25m is 30º. Calculate the horizontal distance between the building and the tree.

15 $\sqrt{3}$ m
5 $\sqrt{3}$ m
25 $\sqrt{3}$ m
30 $\sqrt{3}$ m

Solution:

Concept:

The height difference between the building and the tree is the opposite side, while the horizontal distance is the adjacent side of a right-angled triangle.

Explanation:

Height difference

= 25 - 10 = 15

m.

The angle of elevation is

30^\circ

.

Using

\tan 30^\circ = \frac{\text{opposite}}{\text{adjacent}}

:

\tan 30^\circ = \frac{15}{x}

Since

\tan 30^\circ = \frac{1}{\sqrt{3}}

, we have:

\frac{1}{\sqrt{3}} = \frac{15}{x}

Cross-multiplying gives:

x = 15\sqrt{3}

m

Answer:

The horizontal distance between the building and the tree is

15\sqrt{3}

m.

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