SSC CHSL 11 Aug 2021 Shift 2 Paper

Section: Quantitative Aptitude

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Question : 52 of 100

Marks: +1, -0

Solve for

θ

, if

3 cosec θ+4 \sin θ

-4 √{3}=0

, where

θ

is an acute angle.

$45^{∘}$
$30^{∘}$
$60^{∘}$
$15^{∘}$

Solution:

Given,

3 cosec θ+4 \sin θ-4 √{3}=0

⇒ \;{3}/{\sin θ}+4 \sin θ-4 √{3}=0

{[ ∵ cosec θ=\;{1}/{\sin θ}]}

⇒ 4 \sin ^{2} θ-4 √{3} \sin θ+3=0

Let

\sin θ=x

,

∴ \;\; 4 x^{2}-4 √{3} x+3=0

⇒ \;\; 4 x^{2}-2 √{3} x-2 √{3} x+3=0

⇒ 2 x(2 x-√{3})-√{3}(2 x-√{3})=0

⇒ \;\; (2 x-√{3})(2 x-√{3})=0

∴ \;\; x=\;{√{3}}/{2}

and

\;{√{3}}/{2} ⇒ \sin θ=\;{√{3}}/{2}

Since,

θ

is an acute angle,

\sin θ=\sin 60^{∘}

⇒ \;\; θ=60^{∘}

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