SSC CHSL 13 Apr 2021 Shift 3 Paper

Section: Quantitative Aptitude

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Question : 57 of 100

Marks: +1, -0

Solve the following equation.

θ: 2 \cos ^{2} θ+(4+√{3})

\sin θ-2(1+√{3})=0

where

θ

is an acute angle.

$30^{∘}$
$45^{∘}$
$15^{∘}$
$60^{∘}$

Solution:

Given,

θ: 2 \cos ^{2} θ+(4+√{3})

\sin θ-2(1+√{3})=0

⇒ \;\; 2-2 \sin ^{2} θ+4 \sin θ+√{3} \sin θ

-2-2 √{3}=0

[ ∵ \cos ^{2} θ=1-\sin ^{2} θ]

⇒ 2 \sin ^{2} θ-4 \sin θ-√{3} \sin θ+2 √{3}=0

⇒ \;\; 2 \sin θ(\sin θ-2)-√{3}(\sin θ-2)=0

⇒ \;\; (\sin θ-2)(2 \sin θ-√{3})=0

∴ \;\; \sin θ=2, \;{√{3}}/{2}

Since

θ

is an acute angle,

∴ \;\; Sin θ=\;{√{3}}/{2}=\sin 60^{∘}

θ=60^{∘}

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