The angles of depression of the top and bottom of an 8 { m} tall building from the top of a multi-storied building are 30^{} and 45^{}, respectively. What is the height of the multi-storied building and the distance between the two buildings?

Correct Answer is : 4(3+3) m,4(3+3) m4(3+{3}) { m} , 4(3+{3}) { m}4(3+3) m,4(3+3) m

Correct Answer is : 4(3+3) m,4(3+3) m4(3+{3}) { m} , 4(3+{3}) { m}4(3+3) m,4(3+3) m

Test Index

NIMCET 2025 Paper

Section: Mathematics

Question : 4 of 120

Marks: +1, -0

8 \text{ m}

30^{\circ}

45^{\circ}

Solution:

Concept:

Use angles of depression and trigonometric ratios to relate heights and distance.

Formula:

\tan \theta = \frac{\text{opposite}}{\text{adjacent}}

Solution:

Let height of multi-storied building be

H

m and distance between buildings be

d

Angle of depression to bottom:

\tan 45^\circ = \frac{H}{d} = 1 ⇒ d = H

Angle of depression to top:

\tan 30^\circ = \frac{H - 8}{d} = \frac{1}{\sqrt{3}}

Substitute

d = H

\frac{H - 8}{H} = \frac{1}{\sqrt{3}} ⇒ \sqrt{3}(H - 8) = H

Solve:

\sqrt{3}H - 8\sqrt{3} = H ⇒ H(\sqrt{3} - 1) = 8\sqrt{3} ⇒ H = \frac{8\sqrt{3}}{\sqrt{3} - 1}

Rationalize:

H = \frac{8\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{8\sqrt{3}(\sqrt{3}+1)}{2} = 4\sqrt{3}(\sqrt{3}+1) = 4(3+\sqrt{3})

Thus

d = H = 4(3+\sqrt{3})

Shortcut:

Omitted (no alternate method needed).

Answer:

Option B:

4(3+\sqrt{3}) \text{ m} , 4(3+\sqrt{3}) \text{ m}

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