# A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/6, the angle between sides B and C is (7pi)/12, and the length of B is 11, what is the area of the triangle?

Feb 13, 2016

Find all 3 sides through the use of law of sines, then use Heron's formula to find the Area.

$A r e a = 41.322$

#### Explanation:

The sum of angles:

hat(AB)+hat(BC)+hat(AC)=π

π/6-(7π)/12+hat(AC)=π

hat(AC)=π-π/6-(7π)/12

hat(AC)=(12π-2π-7π)/12

hat(AC)=(3π)/12

hat(AC)=π/4

Law of sines

$\frac{A}{\sin} \left(\hat{B C}\right) = \frac{B}{\sin} \left(\hat{A C}\right) = \frac{C}{\sin} \left(\hat{A B}\right)$

So you can find sides $A$ and $C$

Side A

$\frac{A}{\sin} \left(\hat{B C}\right) = \frac{B}{\sin} \left(\hat{A C}\right)$

$A = \frac{B}{\sin} \left(\hat{A C}\right) \cdot \sin \left(\hat{B C}\right)$

A=11/sin(π/4)*sin((7π)/12)

$A = 15.026$

Side C

$\frac{B}{\sin} \left(\hat{A C}\right) = \frac{C}{\sin} \left(\hat{A B}\right)$

$C = \frac{B}{\sin} \left(\hat{A C}\right) \cdot \sin \left(\hat{A B}\right)$

C=11/sin(π/4)*sin(π/6)

$C = \frac{11}{\frac{\sqrt{2}}{2}} \cdot \frac{1}{2}$

$C = \frac{11}{\sqrt{2}}$

$C = 7.778$

Area

From Heron's formula:

$s = \frac{A + B + C}{2}$

$s = \frac{15.026 + 11 + 7 , 778}{2}$

$s = 16.902$

$A r e a = \sqrt{s \left(s - A\right) \left(s - B\right) \left(s - C\right)}$

$A r e a = \sqrt{16.902 \cdot \left(16.902 - 15.026\right) \left(16.902 - 11\right) \left(16.902 - 7.778\right)}$

$A r e a = 41.322$