# It is the question about THE DIAGRAM. It in B?

Jul 17, 2018

If we use trig to find $S \hat{Q} R$ first

$\cos \theta = \frac{\sqrt{2}}{2}$

$\theta = {\cos}^{-} 1 \left(\frac{\sqrt{2}}{2}\right)$

$S \hat{Q} R = 45$

PQR is a straight line so subtract $S \hat{Q} R$ from 180

$P \hat{Q} S = 135$

Jul 17, 2018

$\angle P Q S$ = ${135}^{0}$

#### Explanation:

Angle $\angle P Q S$ = 180 - Angle $\angle S Q R$

To find $\angle S Q R$ we use SOHCAHTOA.

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

So angle $\angle S Q R$, $\theta$ = $\cos \theta$ = Adjecent/Hypotenuse

$\cos \theta$ = $\frac{\sqrt{2}}{2}$ = $0.701$

$\theta$ = ${\cos}^{- 1} 0.701$ = ${45}^{0}$

Hence, $\angle P Q S$ = $180 - 45$ = ${135}^{0}$