Question

In an acute-angled triangle $ABC$, the points $D,E$and $F$ are the feet of the perpendiculars from $A,B$ and $C$ onto $BC,AC$ and $AB,$ respectively. The point $H$ is the orthocenter. If $\sin A=3/5$ and $BC=39$ then find the length of $AH$

• A. $52$
• B. $65$
• C. $43$
• D. $27$

Answer 1

The correct option is A $52$

Given $\sin A=3/5\Rightarrow \cos A=4/5$

Also $a=39$

$\therefore \frac{a}{\sin A}=2R$

$\frac{39\times 5}{3}=2R$

$2R=65$

$\Rightarrow AH=2R\cos A=65\times \frac{4}{5}=52$