A triangle has two corners with angles of # pi / 6 # 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  /  6  # 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

#0.5624unit^2#

For the largest triangle, the shortest length is reflect to smallest angle.

The angles of triangle are #pi/6= (4pi)/24, (5pi)/8=(15pi)/24, and (5pi)/24#.

Therefor the length side for 1 is #pi/6#

Let another length of triangle are A and B.

Find length of A and B. #A/sin((5pi)/8) = 1/sin(pi/6)# #A = 1/sin(pi/6)*sin((5pi)/8)# #A=1.8478#

#B/sin((5pi)/24) = 1/sin(pi/6)# #B = 1/sin(pi/6)*sin((5pi)/24)# #B=1.2175#

Therefor the largest area for triangle =#1/2ABsin(pi/6)# #=1/2(1.8478)(1.2175)sin(pi/6)# #=1/2(1.8478)(1.2175)(0.5)#

#0.5624unit^2#

Ella Armstrong · Answer

undefined To find the largest possible area of the triangle, we can use the formula for the area of a triangle:

$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$

Since we are given one side of the triangle with a length of 1, and we want to maximize the area, we need to maximize the height of the triangle.

Let's denote the side opposite the angle $ \frac{\pi}{6} $ as $ a $ and the side opposite the angle $ \frac{5\pi}{8} $ as $ b $.

Using the law of sines, we can relate the side lengths to the angles:

$$ \frac{a}{\sin\left(\frac{\pi}{6}\right)} = \frac{1}{\sin\left(\frac{5\pi}{8}\right)} $$

Solving for $ a $:

$$ a = \frac{\sin\left(\frac{\pi}{6}\right)}{\sin\left(\frac{5\pi}{8}\right)} $$

$$ a = \frac{\frac{1}{2}}{\sin\left(\frac{5\pi}{8}\right)} $$

$$ a = \frac{1}{2\sin\left(\frac{5\pi}{8}\right)} $$

Similarly, we can find $ b $ using the same approach:

$$ b = \frac{\sin\left(\frac{5\pi}{8}\right)}{\sin\left(\frac{\pi}{6}\right)} $$

$$ b = \frac{\sin\left(\frac{5\pi}{8}\right)}{\frac{1}{2}} $$

$$ b = 2\sin\left(\frac{5\pi}{8}\right) $$

Now, to maximize the area, we need to maximize the height. The height of the triangle corresponds to the length of $ b $.

Thus, the largest possible area of the triangle is:

$$ \text{Area} = \frac{1}{2} \times 1 \times b $$

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

$$ \text{Area} = \sin\left(\frac{5\pi}{8}\right) $$

Therefore, the largest possible area of the triangle is $ \sin\left(\frac{5\pi}{8}\right) $.