A triangle has two corners with angles of # (2 pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #4 #, what is the largest possible area of the triangle?

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Answer 1 · 2016-11-16T17:35:06

#Area = 4sqrt(3)#

Let #angle A = pi/6#

Let #angle B = (2pi)/3#

Then #angle C = pi - angle A - angle B#

#angle C = pi/6#

Let side #a = 4#

Because #angle A = angle C#, then side #c = 4#, too.

We can use the Law of Cosines to find the length of side b:

#b = sqrt(a^2 + c^2 - 2(a)(c)cos(B))#

#b = sqrt(4^2 + 4^2 - 2(4)(4)cos((2pi)/3))#

#b = 4sqrt(3)#

Let b = the base of the triangle

Let h = the height of the triangle

#h = asin(A)#

#h = 4sin(pi/6)#

#h = 2#

#Area = (1/2)bh#

#Area = (1/2)(4sqrt(3))(2)#

#Area = 4sqrt(3)#