Question
Download Solution PDFदिलेल्या आकृतीमध्ये, ∠BOQ = 60° आणि AB हा वर्तुळाचा व्यास आहे. तर ∠ABO शोधा.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFप्रमेय वापरून, अर्धवर्तुळातील कोन काटकोन असतो,
⇒ ∠BOA = 90°
प्रमेय: पर्यायी खंड प्रमेय असे सांगते की स्पर्शिका आणि जीवा यांच्यातील संपर्क बिंदूद्वारे असलेला कोन पर्यायी विभागातील कोनाइतका असतो.
⇒ ∠BOQ = ∠BAO = 60°
ΔABO मध्ये,
त्रिकोणाच्या कोनांची बेरीज 180° असते
⇒ ∠ABO = 180° – ∠BOA – ∠BAO = 180° – 90° – 60° = 30°
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