In triangle ABC, AB=AC=x, A B C=θ and the circumradius is equal to y. Then x/y equals

Correct Answer is : 2sin⁡θ2 θ2sinθ

Correct Answer is : 2sin⁡θ2 θ2sinθ

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Question : 22 of 90

Marks: +1, -0

\angle A B C=\theta

\frac{x}{y}

equals

Solution:

In triangle ABC, AB=AC=x, therefore

\angle A B C=\angle A C B=\theta

, since angles opposite to equal sides are equal. Let AD be the perpendicular to BC.

\begin{aligned} & \cos \theta=\frac{\text { base }}{\text { hypoteneuse }}=\frac{B D}{A B}=\frac{B D}{x} \Rightarrow B D=x \cos \theta \\ & \cos \theta=\frac{\text { base }}{\text { hypoteneuse }}=\frac{C D}{A B}=\frac{C D}{x} \Rightarrow C D=x \cos \theta \\ & \Rightarrow B C=B D+D C=2 x \cos \theta \\ & \sin \theta=\frac{\text { perpendicular }}{\text { hypoteneuse }}=\frac{A D}{A B}=\frac{A D}{x} \Rightarrow A D=x \sin \theta \end{aligned}

It is given that the circumradius of the triangle is y.

We also know that

\Rightarrow

Area of

\triangle=\frac{a b c}{4 R}

, where R is the circumradius, and a, b, and c are the sides of the triangle.

\begin{aligned} & \Rightarrow \frac{1}{2} \times 2 x \cos \theta \times x \sin \theta=\frac{x \times x \times 2 x \cos \theta}{4 y} \\ & \Rightarrow \frac{1}{2} \times 2 x^{2} \cos \theta \sin \theta=\frac{2 x^{3} \cos \theta}{4 y} \end{aligned}

\begin{aligned} & \Rightarrow \frac{1}{2} \times \sin \theta=\frac{x}{4 y} \\ & \Rightarrow \frac{x}{y}=2 \sin \theta \end{aligned}

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