Acute triangle altitudes and angle between them is given. Find side.

Question

Acute triangle ABC has two altitudes: AD = 5 and CE = 3 the angle between these two altitudes is 60°. Find the AC.

I was trying to find a solution to this problem on internet, but I couldn't find anything.

Any help is highly appreciated!

Answer 1

Let the altitudes intersect at $H$ the orthocenter. I assume that $\angle AHC = 120^{\circ}$, as $\angle DHC = 120^{\circ}$ would mean the triangle is obtuse.

Anyway it's easy to notice that $\angle DHC = \angle EHA = \angle ABC = 60^{\circ}$. Now from the right-angled $\triangle ADB$ we have that: $BA = \frac{AD}{\cos 60^{\circ}} = \frac{10}{\sqrt{3}}$. Similarly $BC = \frac{6}{\sqrt{3}}$. Now from the Cosine Rule for $\triangle ABC$ we have

$$AC = \sqrt{BA^2 + CA^2 - 2\cdot BA \cdot CA \cos \angle ABC} = \sqrt{\frac{100}{3} + \frac{36}{3} - \frac{120}{6}} = \sqrt\frac{76}{3}$$

Answer 2

$$\angle DHC=60^{\circ}=\angle EHA$$ Then $HD:HC=1:2$ and $EH:AH=1:2$

Let $HD=x$ then $HC=2x$ and $EH=y$ then $HA=2y$

Hence, $$ x+2y=5 \\ y+2x=3 $$ $$x=\frac13; y=\frac73$$

$$AE^2=AH^2-EH^2=(2y)^2-(y)^2=3y^2=\frac{49}{3}$$

and

$$AC^2=AE^2+EC^2=\frac{49}{3}+9=\frac{76}{3}$$

Then $$AC=\sqrt{\frac{76}{3}}$$