A geometry problem by Housam Kak

Applying Pythagoras in any triangle we have $BC^2=AC^2+AB^2-2AB\times AC\times \cos A$ implies that $BC= \sqrt{63}$

Using the rule $AB\times AC\times \cos(90-60)^\circ=AH\times BC$ SOLVE IT.

BC^2=AB^2+AC^2-2 AB AC COS(60)=81+36-2 9 6 0.5=63 BC=3radical7 Let HC=x AH^2=AC^2-HC^2=36-x^2 AH^2=AB^2-HB^2=81-(3rad7-x)^2=18+(6rad7)x-x^2 Then 36-x^2=18+(6rad7)x-x^2 So x=(3rad7)/7;x^2=9/7 AH^2=36-9/7=243/7 AH=(9rad21)/7=5.8...=6

Area of triangle ABC is $\frac{1}{2}ab\sin60^\circ=\frac{27\sqrt{3}}{2}$ $BC = \sqrt{9^2+6^2-2\times9\times6\times\cos60^\circ}=\sqrt{63}$

We know that $BC\times h\div2=\frac{27\sqrt{3}}{2}$ .Then we can solve to get 6.

Using cosine rule, we have:

$\begin{aligned} BC^2 & = AB^2 + AC^2 - 2(AB)(AC)\cos \angle BAC \\ & = 9^2 + 6^2 - 2(9)(6) \cos 60^\circ \\ & = 81 + 36 - 2(9)(6) \left( \frac{1}{2}\right) \\ & = 63 \\ \Rightarrow BC & = 3\sqrt{7} \end{aligned}$

Using sine rule, we have:

$\begin{aligned} \frac{\sin \angle ABC}{AC} & = \frac{\sin \angle BAC}{BC} \\ \frac{\sin \angle BAC}{6} & = \frac{\sin 60^\circ}{3\sqrt{7}} \\ \sin \angle BAC & = \frac{6 \sqrt{3}}{2(3 \sqrt{7})} = \sqrt{\frac{3}{7}} \end{aligned}$

Now, we have $[AH] = AB \sin \angle ABC = 9 \times \sqrt{\frac{3}{7}} = 5.8919 \approx \boxed{6}$