જો ત્રિકોણના બે ખૂણાઓનું sine મુલ્ય અનુક્રમે | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $\Delta$,

$\angle A+\angle B+\angle C=180^{\circ}$

$\Rightarrow \quad \angle C=180^{\circ}-(\angle A+\angle B)$

$\Rightarrow \cos C=\cos (180-(A+B))$

$\Rightarrow \cos C=-\cos (A+B)$

${[\cos (\pi-	heta)=-\cos 	heta] }$

$\Rightarrow \cos C=-[\cos A \cos B-\sin A \cdot \sin B]$

$\Rightarrow \cos C=-\left[\frac{12}{13} 	imes \frac{20}{101}-\frac{5}{13} 	imes \frac{99}{101}ight]$

$\Rightarrow \cos C=\frac{255}{1313}$

જો ત્રિકોણના બે ખૂણાઓનું sine મુલ્ય અનુક્રમે $\frac{5}{{13}}$ & $\frac{{99}}{{101}}$ હોય તો ત્રીજા ખૂણાનું cosine મુલ્ય ........... થાય

Similar Questions

$ \cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ ની કિમંત મેળવો.

$\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}\sin \frac{{9\pi }}{{14}}\sin \frac{{11\pi }}{{14}}\sin \frac{{13\pi }}{{14}} = . . . .$

જો $A + B + C = \frac{{3\pi }}{2},$ તો $\cos 2A + \cos 2B + \cos 2C = $

$2\cos x - \cos 3x - \cos 5x = $

જો $cosA + cosB = cosC,\ sinA + sinB = sinC$ હોય તો સમીકરણ $\frac{{\sin \left( {A + B} \right)}}{{\sin 2C}}$ =