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Solution
In ΔPQR, we have
Ext. ∠PRT = ∠RPQ + ∠RQP[Exterior angle theorem]
= 35° + 70° = 105°
In ∠PQR; ∠P + ∠PQR + ∠PRQ = 180°
[Angle sum property theorem]
35° + 70° + ∠PRQ = 180°
∠PRQ = 180° – 105°
∠PRQ = 75°
Ext. ∠SQR = ∠P + ∠PRQ [Exterior angle theorem]
= 35° + 75° = 110°
Hence, ∠PRQ = 75° and ∠SQR = 110°