SSC CGL 17 Aug 2021 Shift 1 Paper

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Question : 74 of 100

Marks: +1, -0

If

\cos (A-B)=\;{√{3}}/{2}

and

\cot (A+B)=\;{1}/{√{3}}

, where

A-B

and

A+B

are acute angles, then

(2 A-3 B)

is equal to

$45^{∘}$
$30^{∘}$
$15^{∘}$
$60^{∘}$

Solution:

\cos (A-B)=\;{√{3}}/{2}

\cos (A-B) \;\; =\cos 30^{∘}

A-B \;\; =30^{∘}

. . . (i)

\;\text " and "\; \cot (A+B) \;\; =\;{1}/{√{3}}

\cot (A+B) \;\; =\cot 60^{∘}

(A+B) \;\; =60^{∘}

. . . (ii)
Adding Eqs. (i) and (ii),

2 A \;\; =90

A \;\; =45^{∘} \;\text " and "\; B=15^{∘}

\;\text " Now, "\;(2 A-3 B) \;\; =2 × 45^{∘}-3 × 15^{∘}

\;\; =90^{∘}-45^{∘}=45^{∘}

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