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

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2 Answers
Jul 17, 2018

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

Jul 17, 2018

/_PQS = 135^0

Explanation:

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