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#