ix.
viii.
110°
1350
Fig. 8
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Answer:
∠PQT+∠PQR=180
∘
(Straight angle made on line TR at point Q)
110
∘
+∠PQR=180
∘
∠PQR=180
∘
−110
∘
=70
∘
In △PQR
∠PSR=∠PQR+∠PRQ (Exterior angle of triangle is eqal to sum of two interior angles)
135
∘
=70
∘
+∠PRQ
∠PRQ=135
∘
−70
∘
=65
∘
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