JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Easy

1 of 1

Application +4 | -1

A force

{P}

is applied on a block placed on a rough horizontal plane. If

{P}=20 {~kg}

and weight of block is

40 {~kg}

, then coefficient of friction is

View answer and explanation

Correct option: C

0.5

Option C (0.5) is correct because the coefficient of friction is calculated directly from the applied force and the weight of the block when the block is on the verge of moving.

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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)=