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 4, what is the area of the triangle?

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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 4, what is the area of the triangle?

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Answer 1 · 2016-02-27T05:57:09

pl,see below

Explanation:

The angle between sides A and B $= 5 \frac{π}{12}$ $=5pi/12$
The angle between sides C and B $= \frac{π}{12}$ $=pi/12$
The angle between sides C and A $= π - 5 \frac{π}{12} - \frac{π}{12} = \frac{π}{2}$ $=pi -5pi/12-pi/12=pi/2$
hence the triangle is right angled one and B is its hypotenuse.
Therefore side A = $B sin (\frac{π}{12}) = 4 sin (\frac{π}{12})$ $Bsin(pi/12)=4sin(pi/12)$
side C = $B cos (\frac{π}{12}) = 4 cos (\frac{π}{12})$ $Bcos(pi/12)=4cos(pi/12)$
So area $= \frac{1}{2} A C sin (\frac{π}{2}) = \frac{1}{2} \cdot 4 sin (\frac{π}{12}) \cdot 4 cos (\frac{π}{12})$ $= 1/2ACsin(pi/2)=1/2*4sin(pi/12)*4cos(pi/12)$
$= 4 \cdot 2 sin (\frac{π}{12}) \cdot cos (\frac{π}{12})$ $=4*2sin(pi/12)*cos(pi/12)$
$= 4 \cdot sin (2 \frac{π}{12})$ $=4*sin(2pi/12)$
$= 4 \cdot sin (\frac{π}{6})$ $=4*sin(pi/6)$
$= 4 \cdot \frac{1}{2}$ $=4*1/2$ =2 sq unit

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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 4, what is the area of the triangle?

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A triangle has sides A, B, and C. The angle between sides A and B is 5π12(5pi)/12 and the angle between sides B and C is π12pi/12. If side B has a length of 4, what is the area of the triangle?

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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 4, what is the area of the triangle?