JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

In a

\triangle A B C

, the sides

a, b

and

c

are such that they are the roots of

x^3-11 x^2+38 x-40=0

. Then,

\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}

is equal to

View answer and explanation

Correct option: C

\frac{9}{16}

Option C (

\frac{9}{16}

) is correct because by using Vieta's formulas to find the sum and sum of products of the roots, and then applying the cosine rule, the expression simplifies to

\frac{9}{16}

More Questions from TRIGONOMETRY RATIOS

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

, where

\alpha=6 \beta

and

\alpha

being an acute angle, then

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

is equal to The value of

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

is If the

\triangle A B C

is acute angled at

C

, then The value of the expression

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

equals

\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)=