∠BAC = 75º ∠ABC + ∠ACB + ∠BAC = 180° ∠ABC + ∠ACB + 75° = 180° ∠ABC + ∠ACB = 180° - 75° = 105° Let, ∠ABC = ∠PBQ = 70° and ∠ACB = ∠RCQ = 35° So, ∠PQR = 180° - (∠PQB + ∠RQC) = 180° - [(180° - 2∠PBQ) + (180° - 2∠RQC) [∵ BQ = PQ; QC = QR] = 180° - [(180° - 2 × 70°) + (180° - 2 × 35°)] = 180° - (40° + 110°) = 180° - 150° = 30°
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