A triangle has two corners with angles of # (2 pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #8 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  (2 pi ) /  3  # and # ( pi  )/  6   #.  If one side of the triangle has a length of #8  #, what is the largest possible area of the triangle?

Sadie Allen · Accepted Answer

Largest possible area of the triangle is 27.7128

Given are the two angles #(2pi)/3# and #pi/6# and the length 8 The remaining angle: #= pi - (((2pi)/3) + pi/6) = pi/6# I am assuming that length AB (8) is opposite the smallest angle. Using the ASA Area#=(c^2*sin(A)*sin(B))/(2*sin(C)# Area#=( 8^2*sin(pi/6)*sin((2pi)/3))/(2*sin(pi/6))# Area#=27.7128#

Sadie Allen · Answer

undefined To find the largest possible area of the triangle, we can use the formula for the area of a triangle, which is given by $ \frac{1}{2} \times \text{base} \times \text{height} $.

Since we know one side of the triangle has a length of 8, we can consider this side as the base of the triangle.

To find the height, we need to drop a perpendicular from the vertex opposite to the 8-unit side to the base. This forms a right triangle with angles $ \frac{\pi}{3} $ and $ \frac{\pi}{6} $.

Using trigonometric ratios, we can determine the height of the triangle in terms of the base. Since the side opposite to $ \frac{\pi}{3} $ is the height, we'll use the tangent function:

$$ \tan\left(\frac{\pi}{3}\right) = \frac{\text{height}}{8} $$

$$ \text{height} = 8 \tan\left(\frac{\pi}{3}\right) $$

Now, we can use the formula for the area of the triangle:

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

$$ \text{Area} = \frac{1}{2} \times 8 \times 8 \tan\left(\frac{\pi}{3}\right) $$

$$ \text{Area} = 32 \tan\left(\frac{\pi}{3}\right) $$

Since $ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} $, we have:

$$ \text{Area} = 32 \sqrt{3} $$

Therefore, the largest possible area of the triangle is $ 32 \sqrt{3} $ square units.