First slide

Trigonometric Identities

Question

The acute angle of a rhombus whose side is a mean proportion al between its diagonals is

Moderate

A

15°
B

20°
C

30°
D

80°

Solution

Let the length of the diagonals of the rhombus be $a$ and $b$ and $x$ be its side then

$\begin{array}{l} \frac{x}{a} = \frac{b}{x} \Rightarrow x^{2} = a b \\ Now \cos θ = \frac{x^{2} + x^{2} - a^{2}}{2 x \cdot x} = \frac{2 a b - a^{2}}{2 a b} = 1 - \frac{a}{2 b} \\ Similarly \cos (180^{\circ} - θ) = 1 - \frac{b}{2 a} \\ \Rightarrow \frac{a}{2 b} = 1 - \cos θ, \frac{b}{2 a} = 1 + \cos θ \\ \Rightarrow 1 - \cos^{2} θ = \frac{1}{4} \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = 30^{\circ} \end{array}$

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