Find $AH$, where $H$ is the orthocenter of $\Delta ABC$

Question

In an acute-angled triangle $\Delta ABC$ , points $D,E,F$ are feet of the perpendiculars drawn from $A,B,C$ onto $BC,CA,AB$ respectively. Suppose $\sin A = \frac{3}{5}$ and $BC = 39$. Find $AH$, where $H$ is the orthocenter of $\Delta ABC$ .

What I Tried: Here is a picture :-

I am a bit weak at trigonometry, I had learnt the use of $\sin,\cos,\tan$ and all in right-angled triangles. Next I came to know that we have the Law of sines and cosines. The Law of sines tell us :- $$\frac{\sin A}{BC} = \frac{\sin B}{AC} = \frac{\sin C}{AB} = \frac{3}{195}$$

From here, I don't know how to proceed. Some other ideas include that $H$ must be inside $\Delta ABC$, as the triangle is acute-angled. But is that intended?

Can anyone help? Thank You.

Answer 1

See here.

Use formula $$AH = 2R \cos A = BC \cot A $$

$$\cot A = \dfrac{4}{3} $$ $$ AH = 52$$

Summary of proof for $AH=2R\cos A$ in the link:

$BFCH$ as constructed is a parallelogram. Opposite sides are equal hence we have this formula.