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Trig (Math) Solved Questions & Notes (2026) - Apex Rankers

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Trig

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2 Questions in this set
TRIGONOMETRY & HEIGHTS: Use trigonometric functions (tangent, angles of elevation and depression) to determine heights of structures or distances from observed points.
Q. 1 Quantitative Aptitude Test
Difficulty: Medium (1 Mark)
The angle of elevation of the top of a 50-meter high tower from a point on the ground at a distance of $50\sqrt{3}\text{ m}$ from its foot is:
A
15°
B
30°
✓ Correct
C
60°
D
75°
💡 Step-by-Step Explanation & Concept Rationale
Using trigonometry: \[\tan \theta = \frac{\text{Height}}{\text{Distance}} = \frac{50}{50\sqrt{3}} = \frac{1}{\sqrt{3}}\] \[\theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = 30^\circ\] Thus, the correct option is (B).
Q. 2 Quantitative Aptitude Test
Difficulty: Medium (1 Mark)
From the top of a vertical tower, the angles of depression of the top and bottom of a 10-meter tall tree are observed to be $30^\circ$ and $45^\circ$ respectively. What is the height of the tower?
A
$5(\sqrt{3}-1)\text{ m}$
B
$10(\sqrt{3}+1)\text{ m}$
C
$5(2+\sqrt{3})\text{ m}$
D
$5(3+\sqrt{3})\text{ m}$
✓ Correct
💡 Step-by-Step Explanation & Concept Rationale
Let $H$ be the height of the tower and $d$ be the horizontal distance between the tower and tree: \[\tan 45^\circ = \frac{H}{d} = 1 \implies d = H\] Looking at the top of the 10 m tree: \[\tan 30^\circ = \frac{H - 10}{d} = \frac{H - 10}{H} \implies \frac{1}{\sqrt{3}} = \frac{H - 10}{H}\] \[H = \sqrt{3}H - 10\sqrt{3} \implies H(\sqrt{3}-1) = 10\sqrt{3}\] \[H = \frac{10\sqrt{3}}{\sqrt{3}-1} = \frac{10\sqrt{3}(\sqrt{3}+1)}{2} = 5\sqrt{3}(\sqrt{3}+1) = 5(3+\sqrt{3})\text{ m}\] Thus, the correct option is (D).
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