In triangle $ABC,$ $AB = 10,$ $BC = 24,$ and $AC = 26.$ Find the length of the shortest altitude in this triangle.
If you know trig: bash each angle
if you don't: note that it is a Pythagorean triple and you coil dddfine var for the st[hottest length, set eq and solve
We have a right triangle with legs AB and BC and hypotenuse AC
The area of this triangle = (1/2)(10) (24) = 120
The shortest altitude is drawn from the vertex opposite the longest side to the longest side ( B to AC )
This altitude can be found as
Area = (1/2) ( AC) (altitude)
120 = (1/2) (26) ( altitude )
120 = 13 (altitude)
120 / 13 = altitude