RPF Constable practice question
Two towers of equal height stand on opposite sides of a straight road that is 180 m wide. From a point on the road between the two towers, the angles of elevation of the tops of the towers are 30° and 45°. Find the height of each tower.
From a RPF Constable Quantitative Aptitude practice set · Quantitative Aptitude
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Correct answer: (A) 65.88 m
Let the common height of the towers be h m, and let the observation point be x m from the foot of the tower seen at 30°, so it is (180 - x) m from the foot of the tower seen at 45°. Then x = h·cot(30°) and (180 - x) = h·cot(45°). Adding, 180 = h[cot(30°) + cot(45°)], so h = 180 / [cot(30°) + cot(45°)] = 180·tan(30°)·tan(45°) / [tan(30°) + tan(45°)]. Substituting tan(30°) = 0.58 and tan(45°) = 1.00, we get h = 65.88 m. (Check: distance from the 30°-tower is 114.12 m and from the 45°-tower is 65.88 m, and 114.12 + 65.88 = 180 m.)
हिंदी में प्रश्न
एक सीधी सड़क, जिसकी चौड़ाई 180 मीटर है, के दोनों किनारों पर बराबर ऊँचाई के दो टॉवर खड़े हैं। सड़क पर दोनों टॉवरों के बीच स्थित किसी बिंदु से टॉवरों की चोटियों के उन्नयन कोण क्रमशः 30° और 45° हैं। प्रत्येक टॉवर की ऊँचाई ज्ञात कीजिए।
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सही उत्तर: (A) 65.88 m
माना दोनों टॉवरों की समान ऊँचाई h मीटर है, तथा प्रेक्षण बिंदु 30° वाले टॉवर के आधार से x मीटर दूर है, तो यह 45° वाले टॉवर के आधार से (180 - x) मीटर दूर होगा। अतः x = h·cot(30°) तथा (180 - x) = h·cot(45°)। दोनों को जोड़ने पर, 180 = h[cot(30°) + cot(45°)], अतः h = 180 / [cot(30°) + cot(45°)] = 180·tan(30°)·tan(45°) / [tan(30°) + tan(45°)]। मान रखने पर tan(30°) = 0.58 और tan(45°) = 1.00, इसलिए h = 65.88 मीटर। (जाँच: 30° वाले टॉवर से दूरी 114.12 मीटर और 45° वाले टॉवर से दूरी 65.88 मीटर है, तथा 114.12 + 65.88 = 180 मीटर।)
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