Question
Download Solution PDFखाली दिलेल्या आकृतीत O हे वर्तुळाचे केंद्र आहे. ∠OPQ = 35° असल्यास, ∠OQP चे मूल्य शोधा.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFपर्यायी खंड प्रमेयानुसार, स्पर्शबिंदूवर स्पर्शिका आणि जीवा यांच्यामध्ये तयार होणारा कोन वर्तुळाच्या पर्यायी भागामध्ये जीवेद्वारे तयार झालेल्या कोनासमान असतो.
दिलेल्या आकृतीमध्ये,
⇒ ∠APQ = ∠QRP = 35°
आपणास माहीत आहे की, वर्तुळाच्या मध्यभागी कमानीच्या आंतरित कोन हा वर्तुळावरील इतर कोणत्याही बिंदूवर कमानीद्वारे केलेल्या कोनाच्या दुप्पट असतो.
⇒ ∠QOP = 2 × ∠QRP = 2 × 35° = 70°
आता, ΔOQP हे विचारात घेतल्यानंतर,
∵ OQ = OP = वर्तुळाची त्रिज्या
⇒ ΔOQP हा समद्विभुज त्रिकोण आहे.
⇒ ∠OQP = ∠OPQ
∵ त्रिकोणाच्या कोनांची बेरीज = 180°
⇒ ∠QOP + ∠OQP + ∠OPQ = 180°
∴ ∠OQP = (180° – 70°)/2 = 110°/2 = 55°
Last updated on Jul 4, 2025
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