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Answer 1 · 2018-07-17T05:06:16

If we use trig to find $S ˆ Q R$ $ShatQR$ first

$cos θ = \frac{\sqrt{2}}{2}$ $cos theta= sqrt2/2$

$θ = {cos}^{- 1} (\frac{\sqrt{2}}{2})$ $theta = cos^-1(sqrt2/2)$

$S ˆ Q R = 45$ $ShatQR=45$

PQR is a straight line so subtract $S ˆ Q R$ $ShatQR$ from 180

$P ˆ Q S = 135$ $PhatQS= 135$

Answer 2 · 2018-07-17T05:21:11

$∠ P Q S$ $/_PQS$ = $135^{0}$ $135^0$

Angle $∠ P Q S$ $/_PQS$ = 180 - Angle $∠ S Q R$ $/_SQR$

To find $∠ S Q R$ $/_SQR$ we use SOHCAHTOA.

In a given triangle SQR, Hypotenuse is $2 c m$ $2cm$ (QS) and Adjacent is $\sqrt{2} c m$ $sqrt(2) cm$

So angle $∠ S Q R$ $/_SQR$ , $θ$ $theta$ = $cos θ$ $costheta$ = Adjecent/Hypotenuse

$cos θ$ $costheta$ = $\frac{\sqrt{2}}{2}$ $sqrt(2)/2$ = $0.701$ $0.701$

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

Hence, $∠ P Q S$ $/_PQS$ = $180 - 45$ $180 - 45$ = $135^{0}$ $135^0$