A triangle has sides A, B, and C. If the angle between sides A and B is $(pi)/12$ , the angle between sides B and C is $(pi)/2$ , and the length of B is 2, what is the area of the triangle?

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Answer 1 · 2017-05-23T10:54:21

Area of triangle is $0.54 (2dp)$ sq.unit.

The angle between sides $A and B$ is $/_c= pi/12=180/12=15^0$

The angle between sides $B and C$ is $/_a= pi/2=180/2=90^0$

The angle between sides $C and A$ is $/_b= 180-(90+15)=75^0$

$B =2 ;$ Appliying sine law we can find $A$ as $A/sina=B/sinb or A =B* sina/sinb or A= 2* sin 90/sin 75 ~~2.07$

Now we know sides $A~~2.07 and B=2$ and their included angle $/_c = 15^0$

Area of triangle $A_t= (A*B*sinc)/2 ~~ (2.07*2*sin15)/2 ~~ 0.54 (2dp)$ sq.unit [Ans]

A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/12(pi)/12, the angle between sides B and C is (pi)/2(pi)/2, and the length of B is 2, what is the area of the triangle?