Question

A triangle has two corners with angles of `pi / 4` and `(3 pi )/ 8`. If one side of the triangle has a length of `15`, what is the largest possible area of the triangle?

Answer 1

Largest possible area of triangle `A = 135.7995`

Explanation:

Three angles are `pi/4, (3pi)/8, (3pi)/8`
It’s an isosceles triangle with side opposite to vertex with `(pi/4)` will provide the largest area possible as it is the least angle of the three.
`a/sin a = b/ sin b = c/ sin c`
`15/ sin (pi/4) = b / sin ((3pi)/8) = c / sin ((3pi)/8)`
Side `b = c = (15*sin ((3pi)/8))/sin (pi/4)`
Side `b = c = 19.5984`
Height `h = (1/2)*15*tan ((3pi)/8) = 18.1066`
Area of triangle `A = (1/2)*b*h = (1/2)*15*18.1066=135.7995`

Aliter :
Area of triangle `A = (1/2)*b*h`
As it’s an isosceles triangle, altitude bisects the base of the triangle.
`h=(1/2)*b*tan ((3pi)/8)`
Area of triangle `A = (1/2)*15*(1/2)*15*tan((3pi)/8) = 135.7995`