triangle trig problem

**Beachball** · August 21st 2009, 04:57 AM

stuck on this part of the question

**red_dog** · August 21st 2009, 08:44 AM

Solution 1. (using cosine rule)

$c^2=4ab\cdot\frac{b^2+c^2-a^2}{2bc}\cdot\frac{a^2+c^2-b^2}{2ac}\Leftrightarrow a^4+b^4-2a^2b^2=0\Leftrightarrow$

$\Leftrightarrow(a^2-b^2)^2=0\Leftrightarrow a^2=b^2\Leftrightarrow a=b$

Solution 2. (using sine rule)

$4R^2\sin^2C=16R^2\sin A\sin B\cos A\cos B\Leftrightarrow\sin^2C=\sin 2A\sin 2B\Leftrightarrow$

$\Leftrightarrow 1-\cos 2C=\cos 2(A-B)-\cos 2(A+B)\leftrightarrow 1-\cos 2C=\cos 2(A-B)-\cos 2C\Leftrightarrow$

$\Leftrightarrow \cos 2(A_B)=0\Leftrightarrow A=B\Leftrightarrow a=b$

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