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 16, 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 16, what is the area of the triangle?

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Answer 1 · 2017-06-30T07:02:50

The area of the triangle is $= 32 u^{2}$ $=32u^2$

Explanation:

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

$= π - (\frac{5}{12} π + \frac{1}{12} π) = \frac{6}{12} π = \frac{1}{2} π$ $=pi-(5/12pi+1/12pi)=6/12pi=1/2pi$

This is a right angle triangle.

$sin (\frac{1}{2} π) = 1$ $sin(1/2pi)=1$

We apply the sine rule

$\frac{16}{sin (\frac{1}{2} π)} = \frac{A}{sin (\frac{1}{12} π)}$ $16/sin(1/2pi)=A/sin(1/12pi)$

$A = 16 sin (\frac{1}{12} π)$ $A=16sin(1/12pi)$

$\frac{16}{1} = \frac{C}{sin (\frac{5}{12} π)}$ $16/1=C/sin(5/12pi)$

$C = 16 sin (\frac{5}{12} π)$ $C=16sin(5/12pi)$

The area of the triangle is

$= \frac{1}{2} A C sin (\frac{1}{2} π) = \frac{1}{2} \cdot 16 sin (\frac{1}{12} π) \cdot 16 sin (\frac{5}{12} π)$ $=1/2ACsin(1/2pi)=1/2*16sin(1/12pi)*16sin(5/12pi)$

$= 128 \cdot 0.25$ $=128*0.25$

$= 32$ $=32$

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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 16, what is the area of the triangle?

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Explanation:

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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 16, what is the area of the triangle?

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Explanation:

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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 16, what is the area of the triangle?