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
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#