Question

If in a $\Delta ABC,a=(1+\sqrt{3})cm,b=2cm$and $\angle C={60}^{\circ }$ find the other two angles and the third side.

Answer 1

we have , $a=1+\sqrt{3},b=2and\angle C={60}^{\circ }$

Weknow that $c^2=a^2+b^2-2abcosC$

$\Rightarrow c^2=(1+\sqrt{3}{)}^2+(2{)}^2-2\times (1+\sqrt{3})\left(2\right)cos{60}^{\circ }$

$\Rightarrow c^2=1+3+2\sqrt{3}+4-4(\sqrt{3}+1)\times \frac{1}{2}$

$\Rightarrow c^2=8+2\sqrt{3}-2\sqrt{3}-2=6$

$\Rightarrow c=\sqrt{6}$

Using sine rule,

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

$\Rightarrow \frac{1+\sqrt{3}}{sinA}=\frac{2}{sinB}=\frac{\sqrt{6}}{sin{60}^{\circ }}$

$\Rightarrow \frac{2}{sinB}=\frac{\sqrt{6}}{\frac{\sqrt{3}}{2}}\Rightarrow sinB=\frac{1}{\sqrt{2}}$

$B={45}^{\circ }$

We know the sum of all angles of a triangle is ${180}^{\circ }$

$\therefore A+B+C={180}^{\circ }$

$\Rightarrow A={180}^{\circ }-(B+C)$

$\Rightarrow A={180}^{\circ }-({45}^{\circ }+{60}^{\circ })={75}^{\circ }$

Hence$\angle A={75}^{\circ },\angle B={45}^{\circ },and,c=\sqrt{6}$