A triangle has two corners with angles of # pi / 2 # and # (3 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  /  2  # and # (3 pi  )/  8   #.  If one side of the triangle has a length of #1  #, what is the largest possible area of the triangle?

Sophia Alfred · Accepted Answer

Largest possible area of the triangle is 1.2071

Given are the two angles #(3pi)/8# and #pi/2# and the length 1 The remaining angle: #= pi - ((3pi)/8) + pi/2) = (pi)/8# I am assuming that length AB (1) is opposite the smallest angle. Using the ASA Area#=(c^2*sin(A)*sin(B))/(2*sin(C))# Area#=( 1^2*sin((3pi)/8)*sin((pi)/2))/(2*sin(pi/8)# Area#=1.2071#

Allison Allen · Answer

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

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

Given that one side of the triangle has a length of 1, and the triangle has two angles of $ \frac{\pi}{2} $ and $ \frac{3\pi}{8} $, we can determine the height of the triangle using trigonometric functions.

Let's denote the angle $ \frac{3\pi}{8} $ as $ \theta $. Then, the height of the triangle can be calculated as:

$$ \text{height} = \sin(\theta) \times \text{side} $$

After finding the height, we can use it to calculate the area of the triangle using the formula mentioned above. We'll maximize the area by maximizing the height, which occurs when $ \theta = \frac{3\pi}{8} $.