Solution: Given, Height of the building (AD) = 20m Angle of depression ( angle.FEA ) = 30° Angle of elevation (angle.CDE ) = 60° From the figure, ABCD is a rectangle. So, AD = BC = 20m and AB = CD . In right angled ∆CDE tan 60° = $\dfrac{ EB + BC}{CD}$ or, √3 = $\dfrac{ x + 20}{CD} or, √3 CD = x + 20 or, CD = $\dfrac{x +20}{√3}$ In…
Read moreExercise 16.1 Other solutions from this unit: Exercise 16.2 - Area of Triangles General Section 1. Find the unknown lengths of the sides of the following figures. [Refer to your book for the figures] a) Solution : Here, $tan45 ^o = \dfrac{x}{20}$ $or, 1 = \dfrac{x}{20}$ $\therefore x = 20m$ And, $sin 45 ^o = …
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