A triangle has two corners with angles of # (2 pi ) / 3 # and # ( pi )/ 6 #. 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 #  (2 pi ) /  3  # and # ( pi  )/  6   #.  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 0.433

Given are the two angles #(2pi)/3# and #pi/6# and the length 1 The remaining angle: #= pi - ((2pi)/3) + pi/6) = (pi)/6# 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(pi/6)*sin((2pi)/3))/(2*sin(pi/6))# Area#=0.433#

Carson Arnold · Answer

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

Area = 1/2 * base * height

In this case, we know one side of the triangle has a length of 1, and we want to maximize the area. To maximize the area, we need to maximize the height of the triangle.

Given that the angles are (2π)/3 and (π)/6, we can determine that the angle opposite the side of length 1 is (π)/6, because the sum of the angles in a triangle is π radians (180 degrees).

Using trigonometry, we can find the height of the triangle opposite the side of length 1:

height = side * tan(angle)

Substituting the values, we get:

height = 1 * tan(π/6)

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

Area = 1/2 * base * height

Substituting the values, we get:

Area = 1/2 * 1 * (1 * tan(π/6))

We can simplify this expression to find the maximum area of the triangle.