Question

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

Answer 1

`"max area " ~~79.550" units"^2` to 3decimal places

Explanation:

`color(blue)("Always sketch a diagram. It usually helps.")`

As the sum of internal angles of a triangle is always `pi` radians (`180^o`) then if two of the triangles are defined it fixes the ratio of the sides.

`color(brown)("The logic driving the solution approach")`
The largest possible area will be derived from the largest possible triangle. So, if we assigned the given length to the shortest side then we have the largest possible triangle.

Angle A is `pi-(3pi)/4-pi/8 = pi(8/8- 6/8-1/8)=(pi)/8`

So angle A = angle C thus triangle ABC is an Isosceles trinagle

Thus the length of AB is the same as Length BC

As angle A and angle C < `pi/4` it follows that AC is the largest length. Thus the shortest length is AB=BC=15

Thus `1/2 AC = ABcos(pi/8)`

Vertical height is `ABsin(pi/8)`

Thus the maximum area is `ABcos(pi/8)xxABsin(pi/8)`

`=> "max area "= 15^2(cos(pi/8)xxsin(pi/8))`

`~~79.550" units"^2` to 3decimal places