The minute hand of a watch is $1.5 \,cm$ long. How far does its tip move in $40$ minutes? ( Use $\pi=3.14$ ).
In $60$ minutes, the minute hand of a watch completes one revolution. Therefore, in $40$ minutes, the minute hand turns through $\frac{2}{3}$ of a revolution. Therefore, ${\theta = 23 \times {{360}^\circ }}$ or $\frac{4 \pi}{3}$ radian. Hence, the required distance travelled is given by
$l=r \theta=1.5 \times \frac{4 \pi}{3} \,cm =2 \pi \,cm =2 \times 3.14 \,cm =6.28 \,cm$
Find the values of other five trigonometric functions if $\tan x=-\frac{5}{12}, x$ lies in second quadrant.
If $\theta $ and $\phi $ are angles in the $1^{st}$ quadrant such that $\tan \theta = 1/7$ and $\sin \phi = 1/\sqrt {10} $.Then
Find the value of:
$\sin 75^{\circ}$
Find the value of the trigonometric function $\cot \left(-\frac{15 \pi}{4}\right)$
Prove that :
$\cot ^{2} \frac{\pi}{6}+\cos ec \,\frac{5 \pi}{6}+3 \tan ^{2}\, \frac{\pi}{6}=6$