With usual notations in $\triangle$ABC, if $b\cos^2\dfrac{C}{2} + c\cos^2\dfrac{B}{2} = \dfrac{3a}{2}$ then

In a triangle ABC, with usual notations $a = \sqrt{3} + 1$, $b = \sqrt{3} - 1$ and $\angle C = 60^\circ$ then the values of $\angle A$ and $\angle B$ respectively are

In a triangle ABC, with the usual notations, $\angle B = \dfrac{\pi}{3}$, $\angle C = \dfrac{\pi}{4}$. If D divides BC internally in the ratio 1:3, then $\dfrac{\sin \angle BAD}{\sin \angle CAD} =$

Let $P_1$, $P_2$, $P_3$ be the altitudes of a triangle ABC from the vertices A, B, C respectively. If $\triangle$ denotes the area of the triangle and s is the semi-perimeter of the triangle, then $\dfrac{\cos A}{P_1} + \dfrac{\cos B}{P_2} + \dfrac{\cos C}{P_3} =$