Question
Download Solution PDF△PQR এ, PQ এবং PR বাহুকে E এবং F পর্যন্ত প্রসারিত করা হয় যাতে ∠Q এবং ∠R এর বাহ্যিক কোণ দ্বিখণ্ডকটি বিন্দু k তে মিলিত হয়। যদি ∠QPR = 110° হয়, তবে ∠QKR = ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFপ্রদত্ত:
∠QPR = 110°
সূত্র:
ত্রিভুজের তিনটি কোণের সমষ্টি হ'ল 180°
∠QKR = 90° - ∠QPR/2
গণনা:
∠RQK = (180° - ∠PQR)/2
∠QRK = (180° - ∠PRQ)/2
△QKR এ
∠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°
সংক্ষিপ্ত কৌশল :
এখানে K হ'ল ΔPQR- এর প্রাক্তন কেন্দ্র, তাই ∠QKR = 90° - (∠QPR/2)
∠QKR = 90° - 110°/2
∴ ∠QKR = 90° - 55° = 35°
Last updated on Jul 17, 2025
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