Question

In ∆PQR, √Q= 30°, √R= 70° and the bisector of √P meet QR at S. Find,
i)√QPS
ii)√PSQ
iii) √PSR ​

Answer 1

Given   : In ∆PQR, ∠Q= 30°, ∠R= 70° and the bisector of ∠P meet QR at S.

To Find :

i)∠QPS

ii)∠PSQ

iii) ∠PSR ​

Solution:

Sum of angles of a triangle = 180°

=> ∠P + ∠Q + ∠R = 180°

=> ∠P + 30° +70° = 180°

=>  ∠P  =  80°

bisector of ∠P meet QR at S.

=> ∠QPS = ∠RPS = 80°/2 = 40°

Hence ∠QPS = 40°

∠PSQ   is exterior angle of triangle PRS

Exterior angle of triangle = Sum of opposite two interior angles

∠PSQ    =  ∠R +  ∠RPS

=> ∠PSQ    = 70°+  40°

=> ∠PSQ    = 110°

∠PSR =   ∠Q +  ∠QPS

=> ∠PSR    = 30°+  40°

∠PSR =   70°

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