Two corners of a triangle have angles of # (5 pi )/ 8 # and # ( pi ) / 12 #. If one side of the triangle has a length of # 18 #, what is the longest possible perimeter of the triangle?

Question

Two corners of a triangle have angles of #  (5  pi  )/  8  # and #  ( pi ) /  12  #. If one side of the triangle has a length of # 18  #, what is the longest possible perimeter of the triangle?

Maverick Smith · Accepted Answer

Longest possible perimeter is #137.434# As two angles are #(5pi)/8# and #pi/12#, third angle is #pi-(5pi)/8-pi/12=(24pi)/24-(15pi)/24-(2pi)/24=(7pi)/24# the smallest of these angles is #pi/12# Hence, for longest possible perimeter of the triangle, the side with length #18#, will be opposite the angle #pi/12#. Now for other two sides, say #b# and #c#, we can use sine formula, and using it #18/sin(pi/12)=b/sin((5pi)/8)=c/sin((7pi)/24)# or #18/0.2588=b/0.9239=c/0.7933# therefore #b=(18xx0.9239)/0.2588=64.259# and #c=(18xx0.7933)/0.2588=55.175# and perimeter is #64.259+55.175+18=137.434#

Theodore Abad · Answer

undefined To find the longest possible perimeter of the triangle, we need to determine the measure of the third angle and then use it to calculate the lengths of the remaining sides.

Given that two angles of the triangle are $ \frac{5\pi}{8} $ and $ \frac{\pi}{12} $, we can find the measure of the third angle by subtracting the sum of these angles from $ \pi $ radians, which is the total measure of angles in a triangle.

$ \text{Third angle} = \pi - \left( \frac{5\pi}{8} + \frac{\pi}{12} \right) $

$ = \pi - \left( \frac{15\pi}{24} + \frac{2\pi}{24} \right) $

$ = \pi - \frac{17\pi}{24} $

$ = \frac{7\pi}{24} $

Now, we'll use the law of sines to find the lengths of the remaining sides.

The law of sines states that for any triangle with sides $a$, $b$, and $c$, and opposite angles $A$, $B$, and $C$, respectively:

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

Let's denote the length of the side opposite the angle $ \frac{5\pi}{8} $ as $a$, and the length of the side opposite the angle $ \frac{\pi}{12} $ as $b$.

Using the law of sines, we have:

$$ \frac{a}{\sin \left( \frac{5\pi}{8} \right)} = \frac{18}{\sin \left( \frac{7\pi}{24} \right)} $$

$$ \frac{b}{\sin \left( \frac{\pi}{12} \right)} = \frac{18}{\sin \left( \frac{7\pi}{24} \right)} $$

Solving for $a$ and $b$, we get:

$$ a = 18 \times \frac{\sin \left( \frac{5\pi}{8} \right)}{\sin \left( \frac{7\pi}{24} \right)} $$

$$ b = 18 \times \frac{\sin \left( \frac{\pi}{12} \right)}{\sin \left( \frac{7\pi}{24} \right)} $$

We then find the length of the third side, $c$, using the law of sines:

$$ c = 18 \times \frac{\sin \left( \pi - \left( \frac{5\pi}{8} + \frac{\pi}{12} \right) \right)}{\sin \left( \frac{7\pi}{24} \right)} $$

$$ c = 18 \times \frac{\sin \left( \frac{7\pi}{24} \right)}{\sin \left( \frac{7\pi}{24} \right)} $$

$$ c = 18 $$

Now, the perimeter of the triangle is the sum of the lengths of its sides:

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

$$ \text{Perimeter} = 18 \times \left( \frac{\sin \left( \frac{5\pi}{8} \right)}{\sin \left( \frac{7\pi}{24} \right)} + \frac{\sin \left( \frac{\pi}{12} \right)}{\sin \left( \frac{7\pi}{24} \right)} + 1 \right) $$

$$ \text{Perimeter} \approx 18 \times \left( \frac{0.3827}{0.724} + \frac{0.2588}{0.724} + 1 \right) $$

$$ \text{Perimeter} \approx 18 \times \left( 0.5289 + 0.3569 + 1 \right) $$

$$ \text{Perimeter} \approx 18 \times 1.8858 $$

$$ \text{Perimeter} \approx 33.945 $$

Therefore, the longest possible perimeter of the triangle is approximately $33.945$.