Properties Of Triangles Ques 17
If in a triangle $A B C, a=1+\sqrt{3} cm, b=2 cm$ and $\angle C=60^{\circ}$, then find the other two angles and the third side.
(1978, 3M)
Correct Answer: 17.$(c=\sqrt{6}, \angle B=45^{\circ}$ and $\angle A=75^{\circ})$ Solution: $
\begin{array}{ll}
\quad a=1+\sqrt{3}, b=2 \text { and } \angle C=60^{\circ} \\
\text { We have, } \quad c^{2}=a^{2}+b^{2}-2 a b \cos C \\
\Rightarrow \quad c^{2}=(1+\sqrt{3})^{2}+4-2(1+\sqrt{3}) \cdot 2 \cos 60^{\circ} \\
\Rightarrow \quad c^{2}=1+2 \sqrt{3}+3+4-2-2 \sqrt{3} \\
\Rightarrow \quad c^{2}=6 \\
\Rightarrow c=\sqrt{6}
\end{array}
$ Using sine rule, $
\begin{aligned}
& \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\
& \Rightarrow \quad \frac{1+\sqrt{3}}{\sin A}=\frac{2}{\sin B}=\frac{\sqrt{6}}{\sin 60^{\circ}} \\
& \therefore \quad \sin B=\frac{2 \sin 60^{\circ}}{\sqrt{6}}=\frac{2 \times \frac{\sqrt{3}}{2}}{\sqrt{6}}=\frac{1}{\sqrt{2}} \\
& \Rightarrow \quad \angle B=45^{\circ} \\
& \therefore \quad \angle A=180^{\circ}-\left(60^{\circ}+45^{\circ}\right)=75^{\circ}
\end{aligned}
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