Question

A triangle has sides A, B, and C. Sides A and B have lengths of 2 and 6, respectively. The angle between A and C is `(pi)/8` and the angle between B and C is `(3pi)/4`. What is the area of the triangle?

Answer 1

Area `= 6sin(pi/8) = 3sqrt(2-sqrt(2)) approx 2.296`.

Explanation:

We know `a=2`, `b=6`, `B=pi/8`, and `A=(3pi)/4`.

We can calculate `C=pi-(pi/8+(3pi)/4) = pi/8`.

The area of the triangle is `1/2a*b*sin(C)`, so:

Area `= 1/2(2)(6)sin(pi/8) = 6sin(pi/8)`

You can find `sin(pi/8)` using the half-angle formula for sine:

`sin(x/2) = sqrt((1-cos(x))/2)`

So `sin(pi/8)= sqrt((1-cos(pi/4))/2)`

`=sqrt((1-sqrt(2)/2)/2) = sqrt(((2-sqrt(2))/2)/2) = sqrt(2-sqrt(2))/2`

so, back to the area:

Area `= 6sin(pi/8) = 6sqrt(2-sqrt(2))/2 = 3sqrt(2-sqrt(2))`

or Area `approx 2.296`