For the triangle ABC , with usual notations, if the angles $A, B, C$ are in A.P. and $\mathrm{m} \angle \mathrm{A}=30^{\circ}, \mathrm{c}=3$, then the values of a and b are respectively

Question

Aicharya · Accepted Answer

Angles A, B, C are in A.P. $\begin{array}{ll} 	herefore \quad & \angle \mathrm{A}+\angle \mathrm{C}=2 \angle \mathrm{~B} \ & \mathrm{Also}, \angle \mathrm{~A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \ & 2 \angle \mathrm{~B}+\angle \mathrm{B}=180^{\circ} \ 	herefore \quad & \angle \mathrm{B}=60^{\circ} \ & \angle \mathrm{A}=30^{\circ} ...[Given]\ 	herefore \quad & \angle \mathrm{C}=90^{\circ} \end{array}$ Using sine Rule $\begin{array}{ll} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \ 	herefore \quad & \frac{a}{\sin 30}=\frac{b}{\sin 60}=\frac{3}{\sin 90} \ 	herefore \quad & \frac{a}{\frac{1}{2}}=\frac{b}{\frac{\sqrt{3}}{2}}=\frac{3}{1} \ & \Rightarrow 2 a=3, \frac{2 b}{\sqrt{3}}=3\end{array}$ $\begin{aligned} & \Rightarrow \mathrm{a}=\frac{3}{2} \ & \Rightarrow \mathrm{~b}=\frac{3 \sqrt{3}}{2}\end{aligned}$

For the triangle ABC , with usual notations, if the angles $A, B, C$ are in A.P. and $\mathrm{m} \angle…

Solution