Properties of Triangles 1 Question 24
24. 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)
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Solution:
- Given that,
$$ \begin{array}{ll} \quad a=1+\sqrt{3}, b=2 \text { and } \angle C=60^{\circ} \\ \text { We have, } \quad c^{2}=a^{2}+b^{2}-2 a b \cos C \\ \Rightarrow \quad & c^{2}=(1+\sqrt{3})^{2}+4-2(1+\sqrt{3}) \cdot 2 \cos 60^{\circ} \\ \Rightarrow \quad & c^{2}=1+2 \sqrt{3}+3+4-2-2 \sqrt{3} \\ \Rightarrow \quad & c^{2}=6 \\ \Rightarrow & c=\sqrt{6} \end{array} $$
Using sine rule,
$$ \begin{aligned} & \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ & \Rightarrow \quad \frac{1+\sqrt{3}}{\sin A}=\frac{2}{\sin B}=\frac{\sqrt{6}}{\sin 60^{\circ}} \\ & \therefore \quad \sin B=\frac{2 \sin 60^{\circ}}{\sqrt{6}}=\frac{2 \times \frac{\sqrt{3}}{2}}{\sqrt{6}}=\frac{1}{\sqrt{2}} \\ & \Rightarrow \quad B=45^{\circ} \\ & \therefore \quad A=180^{\circ}-\left(60^{\circ}+45^{\circ}\right)=75^{\circ} \end{aligned} $$
