AP Police SI practice question
Two tangents PA and PB are drawn from an external point P to a circle with centre O (radius r), touching the circle at A and B. If ∠APB = 72°, find ∠OAB (the base angle of the isosceles triangle OAB, where OA = OB = r).
From a AP Police SI Quantitative Aptitude practice set · Quantitative Aptitude
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Correct answer: (C) 36°
∠AOB = 180° − ∠APB = 180° − 72° = 108°. Since OA = OB = r, triangle OAB is isosceles, so ∠OAB = ∠OBA = (180° − ∠AOB)/2 = (180° − 108°)/2 = 72°/2 = 36°.
हिंदी में प्रश्न
एक बाह्य बिंदु P से केंद्र O (त्रिज्या r) वाले एक वृत्त पर दो स्पर्श रेखाएँ PA और PB खींची जाती हैं, जो वृत्त को A और B पर स्पर्श करती हैं। यदि ∠APB = 72° है, तो ∠OAB (समद्विबाहु त्रिभुज OAB का आधार कोण, जहाँ OA = OB = r) ज्ञात कीजिए।
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सही उत्तर: (C) 36°
∠AOB = 180° − ∠APB = 180° − 72° = 108°। चूँकि OA = OB = r, त्रिभुज OAB समद्विबाहु है, अतः ∠OAB = ∠OBA = (180° − ∠AOB)/2 = (180° − 108°)/2 = 72°/2 = 36°।
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