In a triangle ABC, if a sin A = b sin B, then | vedclass

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(C) Using the Sine Rule,we know that $\frac{a}{\sin A} = \frac{b}{\sin B} = 2R,$ where $R$ is the circumradius of the triangle.This implies $\sin A =

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$a = b$

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(C) Using the Sine Rule,we know that $\frac{a}{\sin A} = \frac{b}{\sin B} = 2R,$ where $R$ is the circumradius of the triangle.This implies $\sin A = \frac{a}{2R}$ and $\sin B = \frac{b}{2R}.$Substituting these into the given equation $a \sin A = b \sin B$:$a \left( \frac{a}{2R} \right) = b \left( \frac{b}{2R} \right)$$\frac{a^2}{2R} = \frac{b^2}{2R}$$a^2 = b^2$Since $a$ and $b$ are side lengths of a triangle,they must be positive,so $a = b.$Thus,the triangle is isosceles.

In a triangle $ABC,$ if $a \sin A = b \sin B,$ then the nature of the triangle is:

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