A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{4}$ $pi/4$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 5, 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 $\frac{π}{4}$ $pi/4$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 5, what is the area of the triangle?

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Answer 1 · 2017-06-14T08:02:41

The area is $= 2.64 u^{2}$ $=2.64u^2$

Explanation:

The angle between $A$ $A$ and $B$ $B$ is

$θ = π - (\frac{1}{12} π + \frac{1}{4} π) = π - \frac{4}{12} π = \frac{2}{3} π$ $theta=pi-(1/12pi+1/4pi)=pi-4/12pi=2/3pi$

$sin θ = sin (\frac{2}{3} π) = \frac{\sqrt{3}}{2}$ $sintheta=sin(2/3pi)=sqrt3/2$

We apply the sine rule to find $= A$ $=A$

$\frac{A}{sin (\frac{1}{12} π)} = \frac{B}{sin (\frac{2}{3} π)}$ $A/sin(1/12pi)=B/sin(2/3pi)$

$A = B \frac{sin (\frac{1}{12} π)}{sin (\frac{2}{3} π)}$ $A=Bsin(1/12pi)/sin(2/3pi)$

The area of the triangle is

$a = \frac{1}{2} A B sin (\frac{1}{4} π)$ $a=1/2ABsin(1/4pi)$

$a = \frac{1}{2} B \frac{sin (\frac{1}{12} π)}{sin (\frac{2}{3} π)} \cdot B \cdot sin (\frac{1}{4} π)$ $a=1/2Bsin(1/12pi)/sin(2/3pi)*B*sin(1/4pi)$

$= \frac{1}{2} \cdot 5 \cdot 5 \cdot sin (\frac{1}{12} π) \cdot \frac{sin (\frac{π}{4})}{sin (\frac{2}{3} π)}$ $=1/2*5*5*sin(1/12pi)*sin(pi/4)/sin(2/3pi)$

$= 2.64$ $=2.64$

Answer 2 · 2017-06-14T08:22:41

$2.64$ $2.64$ $u n i t^{2}$ $unit^2$

Explanation:

Let say the angle, $a = \frac{π}{12} and c = \frac{π}{4}$ $a = pi/12 and c = pi/4$ . Then angle between C and A, $b$ $b$ $= π - \frac{1}{4} π - \frac{1}{12} π = \frac{8}{12} π = \frac{2}{3} π$ $= pi - 1/4 pi - 1/12 pi = 8/12 pi = 2/3 pi$ .

Area of triangle, $= \frac{1}{2} B C sin a$ $= 1/2 BC sin a$
$= \frac{1}{2} (5) (C) sin (\frac{π}{12})$ $= 1/2 (5)(C) sin (pi/12)$ . $\to i$ $->i$

use, $\frac{B}{sin b} = \frac{C}{sin c}$ $B/sin b = C/sin c$ to find C

$\frac{5}{sin (\frac{2}{3} π)} = \frac{C}{sin (\frac{1}{4} π)}$ $5/(sin (2/3 pi)) = C /(sin (1/4 pi))$

$C = \frac{5}{sin (\frac{2}{3} π)} \cdot sin (\frac{1}{4} π) = 4.08$ $C =5/(sin (2/3 pi)) * sin (1/4 pi) = 4.08$ unit

Area of triangle, plug in $C$ $C$ in $\to i$ $->i$
$= \frac{1}{2} (5) (4.08) sin (\frac{π}{12}) = 2.64$ $= 1/2 (5)(4.08) sin (pi/12) = 2.64$ $u n i t^{2}$ $unit^2$

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{4}$ $pi/4$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 5, what is the area of the triangle?

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Explanation:

Explanation:

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{4}$ $pi/4$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 5, what is the area of the triangle?