A triangle has sides A, B, and C. If the angle between sides A and B is $\frac{7 π}{12}$ $(7pi)/12$ , the angle between sides B and C is $\frac{π}{3}$ $pi/3$ , and the length of B is 1, what is the area of the triangle?

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Answer 1 · 2018-01-23T12:06:43

Area of the triangle is $1.62$ $1.62$ sq.unit.

Angle between Sides $A and B$ $A and B$ is $∠ c = \frac{7 π}{12} = \frac{7 \cdot 180}{12} = 105^{0}$ $/_c= (7pi)/12=(7*180)/12=105^0$

Angle between Sides $B and C$ $B and C$ is $/_a= pi/3=180/3=60^0 :.$

Angle between Sides $C and A$ is $/_b= 180-(105+60)=15^0$

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:

$A/sina = B/sinb=C/sinc ; B=1 :. B/sinb=C/sinc$ or

$1/sin15=C/sin105 or C= 1* (sin105/sin15) ~~ 3.73 (2dp)$

Now we know sides $B=1 , C=3.73$ and their included angle

$/_a = 60^0$ . Area of the triangle is $A_t=(B*C*sina)/2$

$:.A_t=(1*3.73*sin60)/2 ~~ 1.62(2dp)$ sq.unit.

Area of the triangle is $1.62$ sq.unit [Ans]

A triangle has sides A, B, and C. If the angle between sides A and B is (7pi)/127π12(7pi)/12, the angle between sides B and C is pi/3π3pi/3, and the length of B is 1, what is the area of the triangle?