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Answer 1 · 2018-07-17T05:06:16

If we use trig to find $ShatQR$ first

$cos theta= sqrt2/2$

$theta = cos^-1(sqrt2/2)$

$ShatQR=45$

PQR is a straight line so subtract $ShatQR$ from 180

$PhatQS= 135$

Answer 2 · 2018-07-17T05:21:11

$/_PQS$ = $135^0$

Angle $/_PQS$ = 180 - Angle $/_SQR$

To find $/_SQR$ we use SOHCAHTOA.

In a given triangle SQR, Hypotenuse is $2cm$ (QS) and Adjacent is $sqrt(2) cm$

So angle $/_SQR$ , $theta$ = $costheta$ = Adjecent/Hypotenuse

$costheta$ = $sqrt(2)/2$ = $0.701$

$theta$ = $cos^(-1) 0.701$ = $45^0$

Hence, $/_PQS$ = $180 - 45$ = $135^0$