In a - ABC prove that sin A-sin Bsin C-

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Quadratic Formula

In a Δ ABC ...

Question

In a $Δ A B C$ prove that $sin A + sin B > sin C$ .

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Solution

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Δ A B C

sin A + sin B > sin C

\frac{sin (A - B)}{sin A sin B} + \frac{sin (B - C)}{sin B sin C} + \frac{sin (C - A)}{sin C sin A} = 0

In any $Δ A B C$ , prove that.

$a (s i n B - s i n C) + b (s i n C - s i n A) + c (s i n A - s i n B) = 0$

Δ A B C

\frac{a^{2} - b^{2}}{c^{2}} = \frac{sin (A - B)}{sin (A + B)}

$I n a Δ A B C, i f c o s C = \frac{s i n A}{2 s i n B}, prove that the triangle is isosceles.$

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