Find the radius of the circle in which a central angle of $60^{\circ}$ intercepts an arc of length $37.4 \,cm$ ( use $\pi=\frac{22}{7}$ ).
Here $l=37.4\, cm$ and $\theta=60^{\circ}=\frac{60 \pi}{180} radian =\frac{\pi}{3}$
Hence, by $r=\frac{l}{\theta},$ we have
$r=\frac{37.4 \times 3}{\pi}=\frac{37.4 \times 3 \times 7}{22}=35.7 \,cm$
If $p = \frac{{2\sin \,\theta }}{{1 + \cos \theta + \sin \theta }}$, and $q = \frac{{\cos \theta }}{{1 + \sin \theta }},$ then
If $x = \sec \theta + \tan \theta ,$ then $x + \frac{1}{x} = $
In a right angled triangle the hypotenuse is $2 \sqrt 2$ times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are
Find the value of $\cos \left(-1710^{\circ}\right)$.
If $\sin \theta + \cos \theta = 1$, then $\sin \theta \cos \theta = $