In ABC, A and AC = 2BC. Then the value of (where R is circumradius of triangle) is

JEE MainMathematicsEasy

In $\bigtriangleup$ ABC, $\angle C = 2\angle$ A and AC = 2BC.
Then the value of
$\frac{2\left( a^{2} + b^{2} + c^{2} \right)}{R^{2}}$ (where R is
circumradius of triangle) is

Show answer and solution

A + B + C = π
Given C = 2A

\Rightarrow

B = π - 3A
As

0 < C < \pi \Rightarrow 0 < 2A < \pi \Rightarrow 0 < A < \frac{\pi}{2}

By sine rule,

\frac{a}{sinA} = \frac{b}{sinB}

\frac{a}{sinA} = \frac{2a}{sin(\pi - 3A)}

\frac{a}{1} = \frac{2a}{3 - 4\sin^{2}A}

or 3 - 4 sin^2 A = 2
or

\sin^{2}A = \frac{1}{4}

sinA = \frac{1}{2}

\ A = \frac{\pi}{6}

\frac{5\pi}{6}

But 0 < A <

\frac{\pi}{2}

\Rightarrow A = \frac{\pi}{6},\angle B = \frac{\pi}{2}

, and

\angle C = \frac{\pi}{3}

\Rightarrow a^{2} + b^{2} + c^{2} = 4R^{2}\left\lbrack \sin^{2}A + \sin^{2}B + \sin^{2}C \right\rbrack

= 4R^{2}\left\lbrack \sin^{2}\frac{\pi}{6} + \sin^{2}\frac{\pi}{2} + \sin^{2}\frac{\pi}{3} \right\rbrack

= 4R^{2}\left\lbrack \frac{1}{4} + 1 + \frac{3}{4} \right\rbrack = 8R^{2}

\frac{a^{2} + b^{2} + c^{2}}{R^{2}}

= 8

Source: Grass-G / M-23 Mathematics Paper 31 Dec FST.docx

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