The angles of elevation of the top of a temple, from the foot and the top of a building 30 m high, are 60° and 30° respectively. Then height of the temple is
45m
CE is the building = 30 m and AD is the temple By symmetry, BD = CE = 30 m, and DE = BC = y m Let AB = x m Also, ∠AED = 60° and ∠ACB = 30° In ∆ ADE,=>tan (∠AED)=AD/DE => tan60= √3=(x+30)/y => x+30=y√3 ------------(i) In ∆ABC,=>tan(∠ACB)=AB/BC =>tan30=1/√3=x/y =>y=x√3 Substituting it in equation (i), we get: => x + 30 = (x √3) × √3 = 3x =>3x-x=2x=30 => x=30/2=15 m ∴AD = AB + BD = 15 + 30 = 45 meters
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