Question

In the given figure, $ABC$ is a triangle in which $\angle C={90}^{\circ }$ and $AB=5$cm.$D$ is a point on $AB$ such that $AD=3$cm and $\angle ACD={60}^{\circ }.$Find the length of side $AC$

Answer 1

Using $m-n$ theorem,

$(3+2)\cot \angle CDA=2\cot {30}^{\circ }-3\cot {60}^{\circ }$

$\Rightarrow \cot \angle CDA=\frac{\sqrt{3}}{5}$

Now, using sine rule in $△CDA,\frac{AC}{\sin \angle CDA}=\frac{AD}{\sin \angle ACD}$

$\Rightarrow AC=\frac{3}{\sin {60}^{\circ }}\times \frac{5}{\sqrt{28}}$

$=\frac{3}{\frac{\sqrt{3}}{2}}\times \frac{5}{2\sqrt{7}}$

$=\frac{3\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\times \frac{5}{\sqrt{7}}$

$=\frac{5\sqrt{3}}{\sqrt{7}}$

$=5\sqrt{\frac{3}{7}}$cm