Question
Download Solution PDFएक वृत्त त्रिभुज ΔABC के परिगत है। O वृत्त का केंद्र है और CP वृत्त के बिंदु C पर स्पर्शरेखा है। यदि BC, ∠OCP को द्विविभाजित करती है, तो ∠BAC का संपूरक कोण क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFजैसा कि हम जानते हैं,
⇒ ∠OCP = 90° (स्पर्शरेखा)
⇒ BC, ∠OCP को द्विविभाजित करती है
∠OCB = ∠OBC = 45° [OB = OC]
ΔBOC में,
⇒ ∠OCB + ∠OBC + ∠BOC = 180°
⇒ ∠BOC + 45° + 45° = 180°
⇒ ∠BOC = 180° – 90° = 90°
⇒ ∠BAC = 1/2 × ∠BOC = (1/2) × 90° = 45°
जैसा कि हम जानते हैं,
दो सम्पूरक कोणों का योग 180° होता है।
माना ∠BAC का पूरक कोण x है, तो
⇒ x + ∠BAC = 180°
⇒ x = 180° – 45° = 135°
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