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?

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Answer 1 · 2018-01-27T11:49:39

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

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 $approx 2.296$

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(pi)/8 and the angle between B and C is (3pi)/4 (3pi)/4. What is the area of the triangle?