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 $\frac{π}{4}$ $pi/4$ , and the length of B is 7, what is the area of the triangle?

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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 $\frac{π}{4}$ $pi/4$ , and the length of B is 7, what is the area of the triangle?

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Answer 1 · 2017-06-19T10:01:35

The area of the triangle is $= 33.5 u^{2}$ $=33.5u^2$

Explanation:

The angle between $A$ $A$ and $C$ $C$ is

$= π - (\frac{7}{12} π + \frac{1}{4} π)$ $=pi-(7/12pi+1/4pi)$

$= π - \frac{10}{12} π$ $=pi-10/12pi$

$= \frac{2}{12} π = \frac{1}{6} π$ $=2/12pi=1/6pi$

$B = 7$ $B=7$

We apply the sine rule

$\frac{A}{sin (\frac{1}{4} π)} = \frac{B}{sin (\frac{1}{6} π)}$ $A/sin(1/4pi)=B/sin(1/6pi)$

$A = \frac{B sin (\frac{1}{4} π)}{sin (\frac{1}{6} π)}$ $A=(Bsin(1/4pi))/sin(1/6pi)$

The area of the triangle is

$= \frac{1}{2} \cdot A \cdot B \cdot sin (\frac{7}{12} π)$ $=1/2*A*B*sin(7/12pi)$

$= \frac{1}{2} \cdot B \cdot \frac{B sin (\frac{1}{4} π)}{sin (\frac{1}{6} π)} \cdot sin (\frac{7}{12} π)$ $=1/2*B*(Bsin(1/4pi))/sin(1/6pi)*sin(7/12pi)$

$= \frac{1}{2} \cdot B^{2} \cdot \frac{sin (\frac{1}{4} π)}{sin (\frac{1}{6} π)} \cdot sin (\frac{7}{12} π)$ $=1/2*B^2*(sin(1/4pi))/sin(1/6pi)*sin(7/12pi)$

$= \frac{1}{2} \cdot 49 \cdot 1.366$ $=1/2*49*1.366$

$= 33.5$ $=33.5$

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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 $\frac{π}{4}$ $pi/4$ , and the length of B is 7, what is the area of the triangle?

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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 $\frac{π}{4}$ $pi/4$ , and the length of B is 7, what is the area of the triangle?