Two corners of a triangle have angles of #pi / 12 # and # pi / 3 #. If one side of the triangle has a length of #6 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #pi  /  12  # and #  pi   /  3  #.  If one side of the triangle has a length of #6  #, what is the longest possible perimeter of the triangle?

Jayden Jackson · Accepted Answer

#18+9\sqrt2+6\sqrt3+3\sqrt6# Let in #\Delta ABC#, #\angle A=\pi/12#, #\angle B=\pi/3# hence #\angle C=\pi-\angle A-\angle B# #=\pi-\pi/12-\pi/3# #={7\pi}/12# For maximum perimeter of triangle , we must consider the given side of length #6# is smallest i.e. side #a=6# is opposite to the smallest angle #\angle A=\pi/12# Now, using Sine rule in #\Delta ABC# as follows #\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}# #\frac{6}{\sin (\pi/12)}=\frac{b}{\sin (\pi/3)}=\frac{c}{\sin ({7\pi}/12)}# #b=\frac{6\sin (\pi/3)}{\sin (\pi/12)}# #b=9\sqrt2+3\sqrt6# & #c=\frac{6\sin ({7\pi}/12)}{\sin (\pi/12)}# #c=12+6\sqrt3# hence, the maximum possible perimeter of the # riangle ABC # is given as #a+b+c# #=6+9\sqrt2+3\sqrt6+12+6\sqrt3# #=18+9\sqrt2+6\sqrt3+3\sqrt6#

Noah Atkinson · Answer

undefined To find the longest possible perimeter of the triangle, we need to consider the Law of Sines, which states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant.

Given that two angles of the triangle are $\frac{\pi}{12}$ and $\frac{\pi}{3}$, we can find the third angle by subtracting the sum of the known angles from $\pi$:

$$\text{Third angle} = \pi - \left(\frac{\pi}{12} + \frac{\pi}{3}\right) = \pi - \left(\frac{5\pi}{12}\right) = \frac{7\pi}{12}$$

Now, we use the Law of Sines to find the lengths of the other two sides:

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

Given that $b = 6$ (one side has a length of 6), we can find the lengths of the other sides using the known angles:

For angle $\frac{\pi}{12}$ opposite side $a$:
$$\frac{a}{\sin \left(\frac{\pi}{12}\right)} = \frac{6}{\sin \left(\frac{7\pi}{12}\right)}$$
$$a = 6 \cdot \frac{\sin \left(\frac{\pi}{12}\right)}{\sin \left(\frac{7\pi}{12}\right)}$$

For angle $\frac{\pi}{3}$ opposite side $c$:
$$\frac{c}{\sin \left(\frac{\pi}{3}\right)} = \frac{6}{\sin \left(\frac{7\pi}{12}\right)}$$
$$c = 6 \cdot \frac{\sin \left(\frac{\pi}{3}\right)}{\sin \left(\frac{7\pi}{12}\right)}$$

Now, we can find the perimeter of the triangle:

$$\text{Perimeter} = a + b + c$$

Plug in the values of $a$, $b$, and $c$ and solve for the perimeter.

$$a = 6 \cdot \frac{\sin \left(\frac{\pi}{12}\right)}{\sin \left(\frac{7\pi}{12}\right)}$$
$$c = 6 \cdot \frac{\sin \left(\frac{\pi}{3}\right)}{\sin \left(\frac{7\pi}{12}\right)}$$

$$a \approx 6 \cdot \frac{\sin \left(\frac{\pi}{12}\right)}{\sin \left(\frac{7\pi}{12}\right)} \approx 6.52$$
$$c \approx 6 \cdot \frac{\sin \left(\frac{\pi}{3}\right)}{\sin \left(\frac{7\pi}{12}\right)} \approx 9.62$$

$$ \text{Perimeter} \approx 6 + 6.52 + 9.62 \approx 22.14$$

So, the longest possible perimeter of the triangle is approximately $22.14$ units.