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

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2 Answers
Mark D.
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#

Bhavin B.
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#

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