The sides of a triangle are $\sin \alpha ,\,\cos \alpha $ and $\sqrt {1 + \sin \alpha \cos \alpha } $ for some $0 < \alpha < \frac{\pi }{2}$. Then the greatest angle of the triangle is.....$^o$

Find the principal solutions of the equation $\sin x=\frac{\sqrt{3}}{2}$

Let $A = \left\{ {\theta \,:\,\sin \,\left( \theta  \right) = \tan \,\left( \theta  \right)} \right\}$ and $B = \left\{ {\theta \,:\,\cos \,\left( \theta  \right) = 1} \right\}$ be two sets. Then

The angles $\alpha, \beta, \gamma$ of a triangle satisfy the equations $2 \sin \alpha+3 \cos \beta=3 \sqrt{2}$ and $3 \sin \beta+2 \cos \alpha=1$. Then, angle $\gamma$ equals

The equation $\sqrt 3 \sin x + \cos x = 4$ has

If  $a = \sin \frac{\pi }{{18}}\sin \frac{{5\pi }}{{18}}\sin \frac{{7\pi }}{{18}}$ and $x$ is the solution of the equatioin $y = 2\left[ x \right] + 2$ and $y = 3\left[ {x - 2} \right] ,$ where $\left[ x \right]$ denotes the integral part of $x,$ then $a$ is equal to :-