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TS EAMCET 18-July-2022 Shift 1 Solved Paper

Section: Mathematics

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Question : 26 of 160

Marks: +1, -0

In a triangle

ABC

(b+c)^2 \sin^2 \frac{A}{2} + (b-c)^2 \cos^2 \frac{A}{2} = K(1-\cos 2A)

K=

$R^2$
$2 R^2$
$R$
$2 R$

Solution: 👈: Video Solution

In

\triangle ABC

,

(b+c)^2 \sin^2 \frac{A}{2} + (b-c)^2 \cos^2 \frac{A}{2} = K(1-\cos 2A)

\Rightarrow (b+c)^2 \sin^2 \frac{A}{2} + (b-c)^2 \left(1 - \sin^2 \frac{A}{2}\right) = K (2 \sin^2 A)

\Rightarrow (b-c)^2 + \{(b+c)^2 - (b-c)^2\} \sin^2 \frac{A}{2} = 2 K \sin^2 A

On putting

\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}} ; \sin A = \frac{a}{2R}

So,

(b-c)^2 + 4bc \left( \frac{(s-b)(s-c)}{bc} \right) = 2K \left( \frac{a^2}{4 R^2} \right)

\Rightarrow (b-c)^2 + 4 \left( \frac{a+b+c}{2} - b \right) \left( \frac{a+b+c}{2} - c \right) = \frac{K a^2}{2 R^2}

\Rightarrow (b-c)^2 + (a+c-b)(a+b-c) = \frac{K a^2}{2 R^2}

\Rightarrow (b-c)^2 + [a - (b-c)][a + (b-c)] = \frac{K a^2}{2 R^2}

\Rightarrow (b-c)^2 + \left[ a^2 - (b-c)^2 \right] = \frac{K a^2}{2 R^2}

\Rightarrow a^2 = \frac{K a^2}{2 R^2} \Rightarrow K = 2 R^2

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