Question

If $\sin {25}^{\circ }.\sin {35}^{\circ }.\sin {85}^{\circ }=\frac{{\mathrm{cos\; x}}^{\circ }}{a}$ where argument of $cos$ is acute and positive then find the value of $(x+a)$.

Answer 1

We have,

$\sin {25}^{\circ }.\sin {35}^{\circ }.\sin {85}^{\circ }=\frac{\cos x}{a}$

$\frac{1}{2}(2\sin {25}^{\circ }.\sin {35}^{\circ }.\sin {85}^{\circ })=\frac{\cos x}{a}$

$\frac{1}{2}\left(\cos \right(-{10}^{\circ })-\cos ({25}^{\circ }+{35}^{\circ }).\sin {85}^{\circ })=\frac{\cos x}{a}$

$\frac{1}{2}\left(\right(\cos {10}^{\circ }-\cos {60}^{\circ }).\sin {85}^{\circ })=\frac{\cos x}{a}$

$\frac{1}{2}(\cos {10}^{\circ }.\sin {85}^{\circ }-\frac{1}{2}.\sin {85}^{\circ })=\frac{\cos x}{a}$

$\frac{1}{2}\left(\frac{1}{2}\right(\sin {95}^{\circ }+\sin {75}^{\circ })-\frac{1}{2}.\sin {85}^{\circ })=\frac{\cos x}{a}$

$\frac{1}{4}\left(\sin \right({90}^{\circ }+5^{\circ })+\sin {75}^{\circ }-\sin ({90}^{\circ }-5^{\circ }\left)\right)=\frac{\cos x}{a}$

$\frac{1}{4}(\cos 5^{\circ }+\sin ({90}^{\circ }-{15}^{\circ })-\cos 5^{\circ })=\frac{\cos x}{a}$

$\frac{1}{4}\cos {15}^{\circ }=\frac{\cos x}{a}$

On comparing both sides, we get

$x=15,a=4$

Since,

$x+a=15+4=19$

Hence, this is the answer.