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

Maverick Smith · Accepted Answer

#=4.732# Clearly this is a right angled triangle with one of the two given angles are #pi/2 and pi/3# and Third angle is #pi-(pi/2+pi/3)=pi-(5pi)/6=pi/6# One #side=hypoten use =2# ;So other sides #=2sin(pi/6) and 2cos( pi/6)# Therefore Perimeter of the triangle#=2+2sin(pi/6)+ 2cos( pi/6)# #=2+(2times0.5)+(2times0.866)# #=2+1+1.732# #=4.732#

Aaliyah Anderson · Answer

undefined To find the longest possible perimeter of the triangle, we can use the Law of Sines.

Let $a$ be the side opposite the angle $\frac{\pi}{3}$, and $b$ be the side opposite the angle $\frac{\pi}{2}$. Since we know the length of one side ($b$), we need to find $a$ and $c$, the side opposite the third angle.

Using the Law of Sines, we have:

$$\frac{a}{\sin\left(\frac{\pi}{3}\right)} = \frac{b}{\sin\left(\frac{\pi}{2}\right)}$$

Given that $b = 2$, and $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\sin\left(\frac{\pi}{2}\right) = 1$, we can solve for $a$:

$$\frac{a}{\frac{\sqrt{3}}{2}} = \frac{2}{1}$$

$$a = \frac{2\sqrt{3}}{3}$$

Now, to find $c$, the side opposite the remaining angle, we can use the fact that the angles in a triangle sum up to $\pi$:

$$\frac{\pi}{3} + \frac{\pi}{2} + \theta = \pi$$

$$\theta = \pi - \frac{\pi}{3} - \frac{\pi}{2}$$

$$\theta = \frac{\pi}{6}$$

Now, we can use the Law of Sines again:

$$\frac{c}{\sin\left(\frac{\pi}{6}\right)} = \frac{2}{\sin\left(\frac{\pi}{2}\right)}$$

$$\frac{c}{\frac{1}{2}} = \frac{2}{1}$$

$$c = 4$$

The perimeter of the triangle is $a + b + c$, so substituting the values we found:

$$P = \frac{2\sqrt{3}}{3} + 2 + 4$$

$$P = \frac{2\sqrt{3}}{3} + 6$$