Question

A triangle has sides A, B, and C. The angle between sides A and B is `(5pi)/12` and the angle between sides B and C is `pi/6`. If side B has a length of 17, what is the area of the triangle?

Answer 1

The area is 144.5.

Explanation:

let's first switch to standard notation. Sides are denotes a, b and c and the angle of the vertex opposite the sides are denotes A, B and C.

The question then becomes, what is the area of the triangle where `b=17`, `A=pi/6` and `C=(5pi)/12`.

The best equation for the areas in `Area=1/2ab sinC`. For this we need the length of side a. To calculate this we also, need angle B.

To calculate angle B we use the fact that `A+B+C=pi`. We rewrite this as `B=pi-A-C`

`B=pi - pi/6-(5pi)/12=(5pi)/12`

Using the sine rule:

`a/(sin A) = b/(sin B)`

This gives:

`a=(b sin A)/(sin B)= (17 sin (pi/6)) /( sin (5pi)/12)`

So the areas is:

`Area = 1/2 (b sin A)/(sin B)= (17 sin (pi/6)) /( sin (5pi)/12) 17 sin (5pi)/12 =17^2 sin(pi/6)`

Now `sin(pi/6)=1/2`

So:

`Area = 17^2/2=289/2=144.5`