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In ∆PQR, ∠Q = 30°, ∠R = 70° and the bisector of ∠P meet QR at S. Find(i) ∠QPS(ii) ∠PSQ(iii) ∠PSR |
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Answer» 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° Learn More: Each exterior angle of a RECTANGLE is - Brainly.in Prove that the angle between the bisectors of two exterior angles of ... |
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