A triangle has two corners with angles of # pi / 12 # and # (5 pi )/ 8 #. If one side of the triangle has a length of #1 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  pi  /  12  # and # (5 pi  )/  8   #.  If one side of the triangle has a length of #1  #, what is the largest possible area of the triangle?

Maverick Smith · Accepted Answer

Area of largest possible triangle #= color (red)(2.0056)#

Three angles are #pi/12, (5pi)/8, (pi - (pi/12) + ((5pi)/8) =( 7pi)/24#

#a/ sin A = b / sin B = c / sin C#

To get the largest possible are, smallest angle should correspond to the side of length 1.

#1 / sin (pi/12) = b / sin ((7pi)/24) = c / sin ((5pi)/8)#

#b = (sin ((7pi)/24)) / (sin (pi/12)# #b = 3.0653#

#c = (sin ((5pi)/8)) / (sin (pi/12))# #c = 3.5696#

Semi perimeter #s = (a + b + c) / 2 = (1+3.0653+3.5696)/2 = 3.8175#

#s-a = 3.8175-1 = 2.8175# #s-b = 3.8175-3.0653 = 0.7522# #s-c = 3.8175-3.5696 = 0.2479#

Area of #Delta = sqrt(s (s-a) (s-b) (s-c))#

#Area of Delta = sqrt(3.8175 * 2.8175 * 0.7522 * 0.2479) = color (red)(2.0056)#

Caroline Aucoin · Answer

undefined To find the largest possible area of the triangle given two angles and one side length, you can use the formula for the area of a triangle:

$$ \text{Area} = \frac{1}{2} \times \text{side}^2 \times \frac{\sin(\text{angle}_1) \times \sin(\text{angle}_2)}{\sin(\text{angle}_1 + \text{angle}_2)} $$

Plugging in the given values:
$$ \text{side} = 1 $$
$$ \text{angle}_1 = \frac{\pi}{12} $$
$$ \text{angle}_2 = \frac{5\pi}{8} $$

$$ \text{Area} = \frac{1}{2} \times 1^2 \times \frac{\sin\left(\frac{\pi}{12}\right) \times \sin\left(\frac{5\pi}{8}\right)}{\sin\left(\frac{\pi}{12} + \frac{5\pi}{8}\right)} $$

$$ \text{Area} = \frac{1}{2} \times \frac{\sin\left(\frac{\pi}{12}\right) \times \sin\left(\frac{5\pi}{8}\right)}{\sin\left(\frac{17\pi}{24}\right)} $$

$$ \text{Area} ≈ 0.35 $$

So, the largest possible area of the triangle is approximately $ 0.35 $ square units.