A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{12}$ $(5pi)/12$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 33, what is the area of the triangle?

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Answer 1 · 2017-05-31T11:11:46

The area of triangle is $136.1 (1 d p)$ $136.1 (1dp)$ sq.unit

The angle between sides $A and B$ $A and B$ is $∠ c = \frac{5 π}{12} = \frac{5 \cdot 180}{12} = 75^{0}$ $/_c = (5pi)/12=(5*180)/12=75^0$

The angle between sides $B and C$ $B and C$ is $∠ a = \frac{π}{12} = \frac{180}{12} = 15^{0}$ $/_a = pi/12=180/12=15^0$

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

Applying sine law we can find $A/sina=B/sinb ; B=33 :. A/sin15=33/sin90 or A= 33*sin 15~~8.54(2 dp); [sin 90 =1]$

Now we know the adjacent sides $A , B$ and their included angle $/_c$ .

The area of triangle is $A_t =(A*B*sin c)/2= (8.54 * 33*sin 75)/2 ~~ 136.1(1dp)$ sq.unit

A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/125π12(5pi)/12 and the angle between sides B and C is pi/12π12pi/12. If side B has a length of 33, what is the area of the triangle?