△PQR में, भुजा PQ और PR को E और F तक इस प्रकार खिंचा गया है कि ∠Q और ∠R के बाहरी कोण द्विभाजक बिंदु k पर मिलते हैं। यदि ∠QPR = 110°, तो ∠QKR = ?

  1. 55°
  2. 45°
  3. 25°
  4. 35°

Answer (Detailed Solution Below)

Option 4 : 35°
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F1Ashish.k 05-10-20 Savita D4

दिया है:

∠QPR = 110°

सूत्र:

एक त्रिभुज के सभी तीन कोणों का योगफल 180° होता है।

∠QKR = 90° - ∠QPR/2

गणना:

∠RQK = (180° - ∠PQR)/2

∠QRK = (180° - ∠PRQ)/2

△QKR में

∠RQK + ∠QRK + ∠QKR = 180°

⇒ [(180° - ∠PQR)/2] + [(180° - ∠PRQ)/2] + ∠QKR = 180°

⇒ 90° - [∠PQR/2] + 90 – [∠PRQ/2] + ∠QKR = 180°

⇒ 180 – [(∠PQR + ∠PRQ)/2] + ∠QKR = 180

⇒ ∠QKR = (∠PQR + ∠PRQ)/2

⇒ 2∠QKR = ∠PQR + ∠PRQ

⇒ 2∠QKR = 180° - ∠QPR

⇒ 2∠QKR = 180° - 110

⇒ ∠QKR = 70/2

∴ ∠QKR = 35°

लघु विधि:

यहाँ, K ΔPQR का वाह्य-केंद्र है, इस प्रकार ∠QKR = 90° - (∠QPR/2)

∠QKR = 90° - 110°/2

∴ ∠QKR = 90° - 55° = 35°

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