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Inscribed Angle Theorem

A point O i...

Question

A point $O$ is the centre of a circle circumscribed about a triangle $A B C$ . Then $¯ ¯¯¯¯¯¯ ¯ O A sin 2 A - ¯ ¯¯¯¯¯¯ ¯ O B sin 2 B + ¯ ¯¯¯¯¯¯ ¯ O C sin 2 C$ is equal to

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Solution

Let $\to V = \to O A s i n A + \to O B s i n 2 B + \to O C s i n 2 C$
$\to V . \to O A = (\to O A) . (\to O A) s i n 2 A + (\to O A) (\to O B) s i n 2 B + (\to O A) . (\to O C) s i n 2 C$
$\Rightarrow \to V . \to O A = R^{2} s i n 2 A + R^{2} c o s 2 C s i n 2 B + R^{2} c o s 2 B s i n 2 C$
$\Rightarrow \to V . \to O A = R^{2} s i n A + R^{2} [s i n 2 B c o s 2 C + c o s 2 B s i n 2 C]$
$= R^{2} s i n 2 A + R^{2} s i n (2 B + 2 C)$
$= R^{2} s i n 2 A - R^{2} s i n 2 A$
$∴ \to V . \to O A = O$
Similarly $\to V . \to O B = O$
$\to V . \to O C = 0$
$∴ \to V . \to O A = \to V . \to O B = \to V . \to O C = O$
Hence, $\to V = 0$
$∴ \to O A s i n 2 A + \to O B s i n 2 B + \to O C s i n 2 C = O$

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