Two corners of a triangle have angles of #pi / 3 # and # pi / 2 #. If one side of the triangle has a length of #9 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #pi  /  3  # and #  pi   /  2  #.  If one side of the triangle has a length of #9  #, what is the longest possible perimeter of the triangle?

Sadie Adams · Accepted Answer

Longest possible perimeter #color(red)(P = 24.59# units

#hat A = pi/3, hat B = pi/2, hat C = pi - pi/3 - pi /2 = pi/6# Side of length 9 should correspond to the least angle #pi/6# to get the longest perimeter. Applying Law of Sines, #a / sin A = b / sin B = c / sin C# #a = (c sin B) / sin C = (9 * sin (pi/6)) / sin (pi/3) = 3 sqrt3# #b = (9 sin(pi/2)) / sin (pi/3) = 6sqrt3# Longest possible perimeter of the triangle is #P = 9 + 6sqrt3 + 3sqrt3 = 24.59# units

Jayden Jackson · Answer

undefined To find the longest possible perimeter of the triangle, we need to maximize the length of the third side.

Using the Law of Sines, we can find the length of the third side. The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant.

So, we have:
$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} $

Given that two angles are $ \frac{\pi}{3} $ and $ \frac{\pi}{2} $, their opposite sides are 9 and the unknown side (let's call it $ c $).

So, we can set up two equations:

For $ \frac{\pi}{3} $:
$ \frac{9}{\sin(\frac{\pi}{3})} = \frac{c}{\sin(C)} $

For $ \frac{\pi}{2} $:
$ \frac{9}{\sin(\frac{\pi}{2})} = \frac{c}{\sin(C)} $

Since $ \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} $ and $ \sin(\frac{\pi}{2}) = 1 $, we can simplify the equations:

For $ \frac{\pi}{3} $:
$ \frac{9}{\frac{\sqrt{3}}{2}} = \frac{c}{\sin(C)} $
$ c = \frac{9}{\frac{\sqrt{3}}{2}} \cdot \sin(C) $

For $ \frac{\pi}{2} $:
$ \frac{9}{1} = \frac{c}{\sin(C)} $
$ c = 9 \cdot \sin(C) $

To maximize the perimeter, we need to maximize the sum of the sides.

Substituting $ c $ from the second equation into the first equation, we get:

$ c = \frac{9}{\frac{\sqrt{3}}{2}} \cdot \sin(C) $
$ c = \frac{18}{\sqrt{3}} \cdot \sin(C) $

Now, the perimeter $ P $ can be expressed as:
$ P = 9 + 9 + c = 18 + \frac{18}{\sqrt{3}} \cdot \sin(C) $

To maximize $ P $, we need to maximize $ \sin(C) $. Since $ \sin(C) $ is maximized when $ C = \frac{\pi}{2} $, the longest possible perimeter occurs when the third angle is $ \frac{\pi}{2} $.

So, the longest possible perimeter of the triangle is:
$ P = 18 + \frac{18}{\sqrt{3}} \cdot \sin(\frac{\pi}{2}) = 18 + \frac{18}{\sqrt{3}} \cdot 1 = 18 + 18\sqrt{3} $