SSC CPO practice question
Three straight lines PQ, RS and EF all pass through the same point O, with rays OR and OE both lying between rays OP and OQ (in the order OP, OR, OE, OQ around point O). If angle POR = (2x + 4)°, angle ROE = (1x + 5)°, and angle EOQ = 123°, find angle SOF.
From a SSC CPO Quantitative Aptitude practice set · Quantitative Aptitude
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Correct answer: (C) 21°
Rays OR and OE both lie between OP and OQ, so angle POR, angle ROE and angle EOQ together make up the straight angle POQ: (2x + 4) + (1x + 5) + 123 = 180. This simplifies to 3x + 132 = 180, so 3x = 48, giving x = 16. So angle ROE = 1(16) + 5 = 21°. Since RS and EF are straight lines through O, ray OS is opposite to OR and ray OF is opposite to OE, so angle SOF is vertically opposite to angle ROE. Therefore angle SOF = angle ROE = 21°.
हिंदी में प्रश्न
तीन सरल रेखाएँ PQ, RS और EF एक ही बिंदु O से होकर गुजरती हैं, जहाँ किरणें OR और OE दोनों किरणों OP और OQ के बीच इस क्रम में स्थित हैं: OP, OR, OE, OQ (बिंदु O के परितः)। यदि कोण POR = (2x + 4)°, कोण ROE = (1x + 5)°, और कोण EOQ = 123° है, तो कोण SOF ज्ञात कीजिए।
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सही उत्तर: (C) 21°
किरणें OR और OE दोनों OP और OQ के बीच स्थित हैं, इसलिए कोण POR, कोण ROE और कोण EOQ मिलकर सीधा कोण POQ बनाते हैं: (2x + 4) + (1x + 5) + 123 = 180। इसे सरल करने पर 3x + 132 = 180, अर्थात 3x = 48, अतः x = 16 प्राप्त होता है। अतः कोण ROE = 1(16) + 5 = 21°। चूँकि RS और EF रेखाएँ O से होकर गुजरने वाली सरल रेखाएँ हैं, किरण OS, OR के विपरीत है और किरण OF, OE के विपरीत है, इसलिए कोण SOF, कोण ROE का शीर्षाभिमुख कोण है। अतः कोण SOF = कोण ROE = 21°।
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