Question

Two corners of a triangle have angles of `(2 pi )/ 3` and `( pi ) / 4`. If one side of the triangle has a length of `15`, what is the longest possible perimeter of the triangle?

Answer 1

`P = 106.17`

Explanation:

By observation, the longest length would be opposite the widest angle, and the shortest length opposite the smallest angle. The smallest angle, given the two stated, is `1/12(pi)`, or `15^o`.

Using the length of 15 as the shortest side, the angles on each side of it are those given. We can calculate the triangle height `h` from those values, and then use that as a side for the two triangular parts to find the other two sides of the original triangle.
`tan(2/3pi) = h/(15-x)` ; `tan(1/4pi) = h/x`

`-1.732 = h/(15-x)` ; `1 = h/x`
`-1.732 xx (15-x) = h` ; AND `x = h` Substitute this for x:

`-1.732 xx (15-h) = h`
`-25.98 + 1.732h = h`

`0.732h = 25.98` ; `h = 35.49`
Now, the other sides are:
`A = 35.49/(sin(pi/4))` and `B = 35.49/(sin(2/3pi))`

`A = 50.19` and `B = 40.98`

Thus, the maximum perimeter is:
`P = 15 + 40.98 + 50.19 = 106.17`

Answer 2

Perimeter`=106.17`

Explanation:

let
`angle A=(2pi)/3`
`angle B=pi/4`
therefore;
using angle sum property
`angle C=pi/12`

Using the sine rule

`a=15×sin ((2pi)/3)/sin (pi/12) = 50.19`
`b=15×(sin ((pi)/4))/sin (pi/12) = 40.98`

perimeter `=40.98+50.19+15 =106.17`