SSC CHSL 12 Oct 2020 Shift 1 Paper

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Question : 69 of 100

Marks: +1, -0

If

\cos (x-y)={\√{3}}/{2}

and

\sin (x+y)={1}/{2}

, then the value of

x(0 < x < 90)

is

$45°$
$30°$
$15°$
$60°$

Solution:

\cos (x-y)={\√{3}}/{2}

\cos (x-y)=\cos 30°

[\∵0°< x < 90°]

x-y=30°

...(i)

\sin (x+y)={1}/{2}

\sin (x+y)=\sin 30°

x+y=30°

...(ii)
Solving Eqs. (i) and (ii), we get

x=30°, y=0°

Hence,

x=30°

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