JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

If in a

\triangle A B C, \sin ^2 A+\sin ^2 B+\sin ^2 C=2

, then the triangle is always

View answer and explanation

Correct option: B

right angled

Option B (right angled) is correct because the given condition

\sin^2 A + \sin^2 B + \sin^2 C = 2

implies that one of the angles must be

90^\circ

, making the triangle right-angled.

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)=