Two corners of a triangle have angles of # (5 pi )/ 8 # and # ( pi ) / 2 #. If one side of the triangle has a length of # 8 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #  (5  pi  )/  8  # and #  ( pi ) /  2  #. If one side of the triangle has a length of # 8  #, what is the longest possible perimeter of the triangle?

Claire Akin · Accepted Answer

Sum needs correction as two angles account for greater than #pi#

Given : /_ A = (5pi)/8, /_B = pi / 2#

Sum of all the three angles must be = #pi#

#pi / 2 + ((5pi)/8) = ((9pi)/8)# which is greater than #pi#

As sum of the given two angles exceeds #pi#, such a triangle cannot exist.

Carson Arnold · Answer

undefined The longest possible perimeter of the triangle can be found by using the Law of Sines to determine the lengths of the other two sides. Let's denote the angles as follows: $A = \frac{5\pi}{8}$, $B = \frac{\pi}{2}$, and $C$ for the remaining angle.

Using the Law of Sines:

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

We know one side ($b$) has a length of 8. Now, we need to find $a$ and $c$.

For $a$:
$$\frac{a}{\sin A} = \frac{8}{\sin \left(\frac{\pi}{2}\right)} = \frac{8}{1} = 8$$
$$a = 8 \sin \left(\frac{5\pi}{8}\right)$$

For $c$:
$$\frac{c}{\sin C} = \frac{8}{\sin \left(\frac{3\pi}{8}\right)}$$
$$c = 8 \sin \left(\frac{3\pi}{8}\right)$$

Now, the perimeter is $a + b + c$.

$$\text{Perimeter} = 8 + 8 \sin \left(\frac{5\pi}{8}\right) + 8 \sin \left(\frac{3\pi}{8}\right)$$

You can calculate the numerical value.