5. In Fig. 6.43, if PQ \bot PS, PQ ||SR, \angle SQR = 28° and \angle QRT = 65°, then find the values of x and y.

                        

Answers (1)

We have, 
PQ \bot PS, PQ || SR, \angle SQR = 28° and \angle QRT = 65°
Now, In \Delta QRS, the side SR produced to T and PQ || RS
therefore, \angleQRT = 28^0+x = 65^0
 So, x = 37^0

Also, \angleQRT = \angleRSQ + \angleSQR (By exterior angle property of a triangle)
Therefore, \angleRSQ = \angleQRT - \angleSQR 
                  \angle RSQ = 65^0-28^0 = 37^0

Now, in \Delta PQS,
\angleP + \anglePQS + \anglePSQ = 180^0
 y=180^0-90^0-37^0
y=52^0

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