A trigo equation

Geometry Level 3

cos ⁡ − 1 41 A = 2 sin ⁡ − 1 3 10 \cos^{-1} \frac{41}{A} = 2\sin^{-1} \frac{3}{10}

Find the value of A A satisfying the above equation.

Hint: you may consider the angle of a sector O P Q OPQ , where the center of the circle is O O with radius 10 and chord P Q = 6 PQ=6 .


The answer is 50.

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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1 solution

Chew-Seong Cheong
Jul 12, 2017

Let θ = cos ⁡ − 1 41 A \theta = \cos^{-1} \dfrac {41}A and ϕ = sin ⁡ − 1 3 10 \phi = \sin^{-1} \dfrac 3{10} . Then we have:

cos ⁡ − 1 41 A = 2 sin ⁡ − 1 3 10 ⟹ θ = 2 ϕ cos ⁡ θ = cos ⁡ 2 ϕ 41 A = 1 − 2 sin ⁡ 2 ϕ = 1 − 2 ( 3 10 ) 2 = 41 50 ⟹ A = 50 \begin{aligned} \cos^{-1} \frac {41}A & = 2 \sin^{-1} \frac 3{10} \\ \implies \theta & = 2\phi \\ \cos \theta & = \cos 2\phi \\ \frac {41}A & = 1 - 2 \sin^2 \phi \\ & = 1 - 2\left(\frac 3{10} \right)^2 \\ & = \frac {41}{50} \\ \implies A & = \boxed{50} \end{aligned}

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