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SSC CHSL 14 Nov 2025 Shift 3 Paper

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Question : 53 of 100

Marks: +1, -0

If

tan\theta =\frac{5}{12}

\theta

is in Quadrant III, what is the value of

sin\theta

?

$\frac{5}{13}$
$-\frac{5}{13}$
$\frac{12}{13}$
$-\frac{12}{13}$

Solution:

Concept:

In Quadrant III, only

\tan\theta

is positive;

\sin\theta

and

\cos\theta

are negative. Using the Pythagorean identity, we find

\sin\theta

from

\tan\theta

.

Explanation:

Given

\tan\theta = \frac{5}{12}

and

\theta

in Quadrant III.

We know

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

. So opposite

=5

, adjacent

=12

.

Hypotenuse

= \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13

.

Thus

\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{5}{13}

in magnitude.

Since

\theta

is in Quadrant III,

\sin\theta

is negative. Therefore

\sin\theta = -\frac{5}{13}

.

Answer:

-\frac{5}{13}

(Option B)

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