A triangle has sides A, B, and C. The angle between sides A and B is (pi)/3 and the angle between sides B and C is pi/6. If side B has a length of 18, what is the area of the triangle?

1 Answer
Rui D.
Dec 31, 2015

The area of the triangle is (81*sqrt(3))/2 or aproximately 70.148

Explanation:

Let's call "c" the angle (=pi/3) between sides A and B, "a" the angle (=pi/6) between sides B and C, and "b" the angle between sides A and C.

For it is a triangle, a+b+c=pi
Then b=pi-c-a=pi-pi/3-pi/6=(6pi-2pi-pi)/6=(3pi) /6=pi/2
(Because of b=pi/2, the triangle is a right one. The side B is the hypotenuse)

In a right triangle it is valid the following statement:
sin x = ("opposed cathetus")/("hypotenuse")

Applying the aforementioned expression
sin a=A/B => A=B*sin a = 18*sin (pi/6) = 18*1/2 = 9
sin c=C/B => C=B*sin c = 18*sin (pi/3) = 18*sqrt(3)/2 = 9*sqrt(3)

Now, to the formula of the triangles area:
S_(triangle) = (base*h_(triangle))/2
(Since the angle between sides A and C is pi/2, it's convenient to use one of this side as the base and the other one as the height of the triangle)

Then
S_(triangle) = (A*C)/2 = :(9*9*sqrt(3))/2 = (81*sqrt(3))/2

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