The lengths of the altitudes of a triangle $ ABC $ are:
$$ h_a = \frac{ 24 \sqrt{ 65}}{ 13 } ~~,~~ h_b = \frac{ 12 \sqrt{ 65}}{ 13 } ~~,~~ h_c = 8 $$The altitude $ h_a $ can be found using formula:
$$ A = \frac{ a \cdot h_a }{2} $$Where $ A $ is a triange area and $ a $ is a length of a side $ BC $. In this eample we have: $ A = 60 $ and $ d(B,C) = \sqrt{ 65 } $ so:
$$ A = \frac{ a \cdot h_a }{2} => h_a = \frac{2 \cdot A}{ a } = \frac{ 2 \cdot 60}{ \sqrt{ 65 } } = \frac{ 24 \sqrt{ 65}}{ 13 } $$