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Ex 9.1, 10 - Two poles of equal heights are standing opposite - Ex 9.1

Ex 9.1, 10 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 2
Ex 9.1, 10 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 3 Ex 9.1, 10 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 4

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Ex 9.1 , 10 Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. Let two poles be AB & CD So, Length of pole = AB = CD Also, Length of the road = 80m So, BC = 80m Lets point P be a point on the road between poles, We need to find height of poles i.e. AB & CD and distance of the point from poles, i.e. BP & CP Since poles are perpendicular to ground ∠ ABP = 90° & ∠ DCP = 90° From (1) & (2) 𝐶𝑃/√3 = √3 BP CP = √3×√3 BP CP = 3BP Now, BC = BP + CP 80 = BP + CP 80 = BP + 3BP 80 = 4BP 4BP = 80 BP = 80/4 BP = 20 m Now, CP = BC – BP CP = 80 – 20 CP = 60 m From (2) CD = √3BP CD = √3 × 20 CD = 20√3 m Hence,Length of the pole = CD = 20√3 m. And Distance of point P from pole CD = CP = 60 m Distance of point P from pole AB = BP = 20 m

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.