In triangle ABC,if (sin A+sin B)(sin A-sin B) | vedclass

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(D) Using the Sine Rule,we have $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$. Substituting $\sin A = ak$,$\sin B = bk$,and $\sin C = c

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(D) Using the Sine Rule,we have $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$. Substituting $\sin A = ak$,$\sin B = bk$,and $\sin C = ck$ into the given equation: $(ak + bk)(ak - bk) = ck(bk + ck)$ $k^2(a^2 - b^2) = k^2(bc + c^2)$ $a^2 - b^2 = bc + c^2$ $a^2 = b^2 + c^2 + bc$ Using the Cosine Rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$. Substituting $a^2 = b^2 + c^2 + bc$: $\cos A = \frac{b^2 + c^2 - (b^2 + c^2 + bc)}{2bc} = \frac{-bc}{2bc} = -\frac{1}{2}$. Since $\cos A = -\frac{1}{2}$,we have $A = 120^{\circ}$.

In $\triangle ABC$,if $(\sin A+\sin B)(\sin A-\sin B)=\sin C(\sin B+\sin C)$,then $\angle A=$ (in $^{\circ}$)

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