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

Question

1503 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-25T12:46:50

$A=(72*sqrt(3))/(1+sqrt(3))$

We use the Formula

$A=1/2*a*b*sin(gamma)$
where $b,gamma$ is given. So we must compute the side length of $a$ :

The third angle is given by $5pi/12$
so we can use the Theorem of sines:

$sin(5/12*pi)/sin(pi/3)=12/a$

From here we get

$a=12sin(pi/3)/sin(5*pi/12)$
putting Things together we get

$A=1/2*12*12*sin(pi/3)*sin(pi/4)/sin(5*pi/12)$

Note that

$sin(pi/3)=sqrt(3)/2$

$sin(5*pi/12)=(1+sqrt(3))/(2*sqrt(2))$

$sin(pi/4)=sqrt(2)/2$
thus

$A=(72*sqrt(3))/(1+sqrt(3))$

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