# 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?

Dec 27, 2017

$P = 106.17$

#### 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 $\frac{1}{12} \left(\pi\right)$, 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.
$\tan \left(\frac{2}{3} \pi\right) = \frac{h}{15 - x}$ ; $\tan \left(\frac{1}{4} \pi\right) = \frac{h}{x}$

$- 1.732 = \frac{h}{15 - x}$ ; $1 = \frac{h}{x}$
$- 1.732 \times \left(15 - x\right) = h$ ; AND $x = h$ Substitute this for x:

$- 1.732 \times \left(15 - h\right) = h$
$- 25.98 + 1.732 h = h$

$0.732 h = 25.98$ ; $h = 35.49$
Now, the other sides are:
$A = \frac{35.49}{\sin \left(\frac{\pi}{4}\right)}$ and $B = \frac{35.49}{\sin \left(\frac{2}{3} \pi\right)}$

$A = 50.19$ and $B = 40.98$

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

Dec 27, 2017

Perimeter$= 106.17$

#### Explanation:

let
$\angle A = \frac{2 \pi}{3}$
$\angle B = \frac{\pi}{4}$
therefore;
using angle sum property
$\angle C = \frac{\pi}{12}$

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 $= 40.98 + 50.19 + 15 = 106.17$