Question

In $\Delta ABC$, if $\frac{AB}{1}=\frac{AC}{2}=\frac{BC}{\sqrt{3}}$, then $m\angle C=.......$

• A. ${90}^o$
• B. ${30}^o$
• C. ${60}^o$
• D. ${45}^o$

Answer 1

The correct option is B ${30}^o$

From the law of cosines,

$A{B}^2=B{C}^2+A{C}^2-2\left(BC\right)\left(AC\right)cosC$

Given that $AC=2AB$ and $BC=\sqrt{3}AB$

Substituting in the cosine equation,

$A{B}^2=3A{B}^2+4A{B}^2-2\left(\sqrt{3}AB\right)\left(2AB\right)cosC$

which implies

$6A{B}^2=4\sqrt{3}A{B}^{2}cosC$

$cosC=\frac{6}{4\sqrt{3}}$

$cosC=\frac{\sqrt{3}}{2}$

therefore $\angle C={30}^0$ (since, $cos{30}^0=\frac{\sqrt{3}}{2}$)