A triangle has sides A, B, and C. The angle between sides A and B is $pi/4$ and the angle between sides B and C is $pi/12$ . If side B has a length of 13, what is the area of the triangle?

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Answer 1 · 2017-11-12T11:02:41

The area of the triangle is $17.86$ sq.unit.

Angle between Sides $A and B$ is $/_c= (pi)/4=180/4=45^0$

Angle between Sides $B and C$ is $/_a= pi/12=180/12=15^0 :.$

Angle between Sides $C and A$ is $/_b= 180-(45+15)=120^0$

The sine rule states if $A, B and C$ are the lengths of the sides

and opposite angles are $a, b and c$ in a triangle, then:

$A/sina = B/sinb=C/sinc ; B=13 :. A/sina=B/sinb$ or

$A/sin15=13/sin120 or A = 13* sin15/sin120 ~~ 3.89(2dp)$ unit.

Now we know $A=3.89 , B=13$ and their included angle

$/_c=45^0$ . The area of triangle is $A_t=(A*B*sinc)/2$

$:. A_t=(3.89*13*sin45)/2 ~~17.86 (2dp)$ sq.unit.

The area of the triangle is $17.86 (2dp)$ sq.unit. [Ans]

A triangle has sides A, B, and C. The angle between sides A and B is pi/4pi/4 and the angle between sides B and C is pi/12pi/12. If side B has a length of 13, what is the area of the triangle?