A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{6}$ $pi/6$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 3, 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 $\frac{π}{6}$ $pi/6$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 3, what is the area of the triangle?

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Answer 1 · 2016-01-08T16:43:41

$A r e a = 0.8235$ $Area=0.8235$ square units.

Explanation:

First of all let me denote the sides with small letters $a$ $a$ , $b$ $b$ and $c$ $c$ .
Let me name the angle between side $a$ $a$ and $b$ $b$ by $∠ C$ $/_ C$ , angle between side $b$ $b$ and $c$ $c$ by $∠ A$ $/_ A$ and angle between side $c$ $c$ and $a$ $a$ by $∠ B$ $/_ B$ .

Note:- the sign $∠$ $/_$ is read as "angle".
We are given with $∠ C$ $/_C$ and $∠ A$ $/_A$ . We can calculate $∠ B$ $/_B$ by using the fact that the sum of any triangles' interior angels is $π$ $pi$ radian.
$\Rightarrow ∠ A + ∠ B + ∠ C = π$ $implies /_A+/_B+/_C=pi$
$\Rightarrow \frac{π}{12} + ∠ B + \frac{π}{6} = π$ $implies pi/12+/_B+(pi)/6=pi$
$\Rightarrow ∠ B = π - (\frac{π}{6} + \frac{π}{12}) = π - \frac{3 π}{12} = π - \frac{π}{4} = \frac{3 π}{4}$ $implies/_B=pi-(pi/6+pi/12)=pi-(3pi)/12=pi-pi/4=(3pi)/4$
$\Rightarrow ∠ B = \frac{3 π}{4}$ $implies /_B=(3pi)/4$

It is given that side $b = 3 .$ $b=3.$

Using Law of Sines
$\frac{sin ∠ B}{b} = \frac{sin ∠ C}{c}$ $(Sin/_B)/b=(sin/_C)/c$
$\Rightarrow \frac{sin (\frac{3 π}{4})}{3} = \frac{sin (\frac{π}{6})}{c}$ $implies (Sin((3pi)/4))/3=sin((pi)/6)/c$

$\Rightarrow \frac{\frac{1}{\sqrt{2}}}{3} = \frac{\frac{1}{2}}{c}$ $implies (1/sqrt2)/3=(1/2)/c$

$\Rightarrow \frac{\sqrt{2}}{6} = \frac{1}{2 c}$ $implies sqrt2/6=1/(2c)$

$\Rightarrow c = \frac{6}{2 \sqrt{2}}$ $implies c=6/(2sqrt2)$

$\Rightarrow c = \frac{3}{\sqrt{2}}$ $implies c=3/sqrt2$

Therefore, side $c = \frac{3}{\sqrt{2}}$ $c=3/sqrt2$

Area is also given by
$A r e a = \frac{1}{2} b c sin ∠ A$ $Area=1/2bcSin/_A$

$\Rightarrow A r e a = \frac{1}{2} \cdot 3 \cdot \frac{3}{\sqrt{2}} sin (\frac{π}{12}) = \frac{9}{2 \sqrt{2}} \cdot 0.2588 = 0.8235$ $implies Area=1/2*3*3/sqrt2Sin((pi)/12)=9/(2sqrt2)*0.2588=0.8235$ square units
$\Rightarrow A r e a = 0.8235$ $implies Area=0.8235$ square units

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{6}$ $pi/6$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 3, 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 $\frac{π}{6}$ $pi/6$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 3, what is the area of the triangle?