Can You Solve This Without Trigonometry?

First we have that $\angle A=60^{o}$ , and the bisector intersects $\overline{BC}$ at $D$ . As $\angle DAB=30^{o} \rightarrow \angle ADB=60^{o}$ this means, if the bisector equals $x$ we have $\frac{x}{2}\sqrt{3}=5 \rightarrow x=\frac{10}{3}\sqrt{3}$

We know that $\angle A=60^\circ$ . Divide this into two equal angles (see my figure). Let $x$ be the length of the angular bisector. Then,

$\large \cos 30^\circ=\dfrac{adjacent~side}{hypotenuse}=\dfrac{5}{x}$ $\implies$ $\large x=\dfrac{5}{\cos~30^\circ} \approx$ $\boxed{5.77}$

Using Trigonometry(CAH): $cos=\dfrac{adj}{hyp}$

Let $AD$ be the length of the angular bisector. Since it is right triangle, $\angle BAC=90-30=60^\circ$ . We divide it by $2$ , we get: $\angle DAC=\angle BAD=30^\circ$ . Then,

$cos~30=\dfrac{5}{AD}$

$AD=\dfrac{5}{cos~30}=5.77$ $\boxed{\text{answer correct to two decimal places}}$