A triangle has sides A, B, and C. The angle between sides A and B is $pi/6$ and the angle between sides B and C is $pi/12$ . If side B has a length of 18, what is the area of the triangle?

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Answer 1 · 2017-08-12T11:56:01

Area of the triangle is $29.65$ sq.unit.

The angle between sides A and B is $/_c = pi/6 =180/6=30^0$

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

The angle between sides C and A is

$/_b = 180-(30+15)=135^0 , B=18$ Applying sine law we get

$A/sina = B/sinb or A= B*sina/sinb= 18 *sin15/sin135 ~~ 6.59$

Now we have side $A = 6.59 , B=18$ and their included angle

$/_c = 30^0$ . Area of the triangle is $A_t=(A*B*sinc)/2$ or

$A_t=(6.59*18*sin30)/2 ~~ 29.65$ sq.unit [Ans]

A triangle has sides A, B, and C. The angle between sides A and B is pi/6pi/6 and the angle between sides B and C is pi/12pi/12. If side B has a length of 18, what is the area of the triangle?