Question

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?

Answer 1

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`