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

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Question

Given,

Let the angles be $A, B \,and\, C$. $\sin A=\frac{5}{13}$ and es $\sin B=\frac{99}{101}$

we know;

$\cos 	heta=\sqrt{1-\sin ^2 	heta}

Vedclass · Accepted Answer

$725/1313$

Vedclass · Answer

Given,

Let the angles be $A, B \,and\, C$. $\sin A=\frac{5}{13}$ and es $\sin B=\frac{99}{101}$

we know;

$\cos 	heta=\sqrt{1-\sin ^2 	heta}$

Hence,

$\cos A =\sqrt{1-\sin ^2 A}=\sqrt{1-\left(\frac{5}{13}ight)^2}$

$=\sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$

simitarly, $\cos B=\sqrt{1-\sin ^2 B}=\frac{20}{101}$.

In a $\Delta$,

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow \quad \angle C=180^{\circ}-(\angle A+\angle B)$

$\Rightarrow \cos C=\cos (180-(A+B))$

$\Rightarrow \cos C=-\cos (A+B)$

${[\cos (\pi-	heta)=-\cos 	heta] }$

$\Rightarrow \cos C=-[\cos A \cos B-\sin A \cdot \sin B]$

$\Rightarrow \cos C=-\left[\frac{12}{13} 	imes \frac{20}{101}-\frac{5}{13} 	imes \frac{99}{101}ight]$

$\Rightarrow \cos C=\frac{255}{1313}$

The sines of two angles of a triangle are equal to $\frac{5}{{13}}$ & $\frac{{99}}{{101}}.$ The cosine of the third angle is :

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