A triangle has sides A,B, and C. If the angle between sides A and B is (pi)/3π3, the angle between sides B and C is pi/6π6, and the length of B is 9, what is the area of the triangle?

1 Answer
Sid P.
Dec 28, 2015

17.537

Explanation:

As the sum of all the angles of a triangle is 180^01800 or piπ radians
So, from your sum it can be seen that the third angle between A and C =pi-pi/3-pi/6=pi/2=π−π3−π6=π2
Thus making the side B, which is opposite to the 90^0900 angle hypotenuse of the triangle.
So, the area of the triangle would be 1/2*base*height=1/2*A*C12⋅base⋅height=12⋅A⋅C
Given that side B=9B=9, we can find the other two side by triangle geometry
A=B*cos(pi/3)=9*cos(pi/3)=4.5A=B⋅cos(π3)=9⋅cos(π3)=4.5
C=B*sin(pi/6)=9*sin(pi/6)=7.794C=B⋅sin(π6)=9⋅sin(π6)=7.794
Hence, Area=1/2*A*C=1/2*4.5*7.794=17.537Area=12⋅A⋅C=12⋅4.5⋅7.794=17.537

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