A triangle has sides A, B, and C. The angle between sides A and B is $(3pi)/4$ and the angle between sides B and C is $pi/12$ . If side B has a length of 27, what is the area of the triangle?

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Answer 1 · 2018-07-15T19:02:46

$A=1/4*27^2*sqrt(2)*(sqrt(3)+1)$

At first we will compute the third angle:
$pi-3/4*pi-pi/12=(12pi-9pi-pi)/12=pi/6$
with the Theorem of sines weget

$a=(27*sin(pi/6))/(sin(pi/12))$
note that

$sin(pi/6)=1/2$

$sin(pi/12)=(sqrt(3)-1)/(2sqrt(2))$
Now we use the Formula
$A=1/2*a*b*sin(gamma)$

so we get

$A=1/2*27^2*sqrt(2)/(sqrt(3)-1)$
this is

$A=1/4*27^2*(sqrt(3)+1)$

We have used that
$1/(sqrt(3)-1)=(sqrt(3)+1)/2$

A triangle has sides A, B, and C. The angle between sides A and B is (3pi)/4(3pi)/4 and the angle between sides B and C is pi/12pi/12. If side B has a length of 27, what is the area of the triangle?