JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

In a

\triangle A B C

, if

a=8, b=15, c=17

, then

\sin \frac{A}{2}

and

\cos A

are equal to

View answer and explanation

Correct option: A

\frac{1}{\sqrt{17}}, \frac{15}{17}

Option A (

\frac{1}{\sqrt{17}}, \frac{15}{17}

) is correct because by using the given side lengths, we can calculate the semi-perimeter and then apply the half-angle formula for

\sin \frac{A}{2}

and the double-angle formula for

\cos A

More Questions from TRIGONOMETRY RATIOS

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\tan \alpha=\frac{p}{q}

\alpha=6 \beta

\alpha

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

\triangle A B C

C

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=