The sines of two angles of a triangle are fra | vedclass

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(B) Let the angles of the triangle be $A, B,$ and $C$. Given $\sin A = \frac{5}{13}$ and $\sin B = \frac{99}{101}$.Since $A$ and $B$ are angles of a t

Vedclass · Accepted Answer

$255$

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(B) Let the angles of the triangle be $A, B,$ and $C$. Given $\sin A = \frac{5}{13}$ and $\sin B = \frac{99}{101}$.Since $A$ and $B$ are angles of a triangle,$\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.Similarly,$\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \frac{9801}{10201}} = \sqrt{\frac{400}{10201}} = \frac{20}{101}$.In a triangle,$A + B + C = 180^{\circ}$,so $C = 180^{\circ} - (A + B)$.Therefore,$\cos C = \cos(180^{\circ} - (A + B)) = -\cos(A + B)$.Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:$\cos C = -(\cos A \cos B - \sin A \sin B) = \sin A \sin B - \cos A \cos B$.Substituting the values:$\cos C = \left(\frac{5}{13} 	imes \frac{99}{101}ight) - \left(\frac{12}{13} 	imes \frac{20}{101}ight) = \frac{495}{1313} - \frac{240}{1313} = \frac{255}{1313}$.

The sines of two angles of a triangle are $\frac{5}{13}$ and $\frac{99}{101}$. The cosine of the third angle is: (in $/1313$)

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