Question

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

• A. $5\sqrt{\frac{3}{7}}cm$
• B. $\sqrt{\frac{7}{3}}cm$
• C. $\frac{3}{\sqrt{7}}cm$
• D. none of these

Answer 1

The correct option is A $5\sqrt{\frac{3}{7}}cm$

Consider the given triangle. $DECF$ is a rectangle

Now, $\angle ADE=\angle BDF=\angle B$

Thus, $DE=3\cos B$

$=CD\sin {60}^0$

Hence, $3\cos B=\frac{\sqrt{3}CD}{2}$

$\cos B=\frac{CD}{2\sqrt{3}}$...(i)

And
$2\sin B=CD\sin {30}^0$

$\sin B=\frac{CD}{4}$ ...(ii)

Taking the ratio of i and ii, we get

$\cot B=\frac{2}{\sqrt{3}}$.

Hence $\sin B=\sqrt{\frac{3}{7}}$ and $\cos B=\frac{2}{\sqrt{7}}$.

Now
$AC=5\sin B$

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