In $\triangle ABC$,if $a, b, c$ are in arithmetic progression and $C = 2A$,then $a: c =$

Let $a, b$ and $c$ denote the lengths of sides $BC, CA$ and $AB$ of $\triangle ABC$. In $\triangle ABC$,$\angle BAC = 30^{\circ}$ and $\angle ABC = 60^{\circ}$. Then $a: b: c$ is

In $\triangle ABC$,$a=6 \text{ cm}$,$b=10 \text{ cm}$ and $c=14 \text{ cm}$. Then,the sum of the acute angles of the triangle is (in $^{\circ}$)

In a triangle,the lengths of the two larger sides are $10 \, cm$ and $9 \, cm$ respectively. If the angles of the triangle are in $A.P.$,then the length of the third side in $cm$ can be

Prove that in any triangle $ABC$,$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$,where $a, b, c$ are the lengths of the sides opposite to the vertices $A, B, C$ respectively.

In $\triangle ABC, (r_2+r_3) \operatorname{cosec}^2 \frac{A}{2} =$