A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{12}$ $pi/12$ , vertex B has an angle of $\frac{3 π}{4}$ $(3pi)/4$ , and the triangle's area is $5$ $5$ . What is the area of the triangle's incircle?

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A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{12}$ $pi/12$ , vertex B has an angle of $\frac{3 π}{4}$ $(3pi)/4$ , and the triangle's area is $5$ $5$ . What is the area of the triangle's incircle?

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Answer 1 · 2017-07-15T16:49:13

The area of the incircle is $= 1.34 u^{2}$ $=1.34u^2$

Explanation:

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The area of the triangle is $A = 5$ $A=5$

The angle $ˆ A = \frac{1}{12} π$ $hatA=1/12pi$

The angle $ˆ B = \frac{3}{4} π$ $hatB=3/4pi$

The angle $ˆ C = π - (\frac{1}{12} π + \frac{3}{4} π) = \frac{1}{6} π$ $hatC=pi-(1/12pi+3/4pi)=1/6pi$

The sine rule is

$\frac{a}{sin ˆ A} = \frac{b}{sin ˆ B} = \frac{c}{sin ˆ C} = k$ $a/(sin hat (A))=b/sin hat (B)=c/sin hat (C)=k$

So,

$a = k sin ˆ A$ $a=ksin hatA$

$b = k sin ˆ B$ $b=ksin hatB$

$c = k sin ˆ C$ $c=ksin hatC$

Let the height of the triangle be $= h$ $=h$ from the vertex $A$ $A$ to the opposite side $B C$ $BC$

The area of the triangle is

$A = \frac{1}{2} a \cdot h$ $A=1/2a*h$

But,

$h = c sin ˆ B$ $h=csin hatB$

So,

$A = \frac{1}{2} k sin ˆ A \cdot c sin ˆ B = \frac{1}{2} k sin ˆ A \cdot k sin ˆ C \cdot sin ˆ B$ $A=1/2ksin hatA*csin hatB=1/2ksin hatA*ksin hatC*sin hatB$

$A = \frac{1}{2} k^{2} \cdot sin ˆ A \cdot sin ˆ B \cdot sin ˆ C$ $A=1/2k^2*sin hatA*sin hatB*sin hatC$

$k^{2} = \frac{2 A}{sin ˆ A \cdot sin ˆ B \cdot sin ˆ C}$ $k^2=(2A)/(sin hatA*sin hatB*sin hatC)$

$k = \sqrt{\frac{2 A}{sin ˆ A \cdot sin ˆ B \cdot sin ˆ C}}$ $k=sqrt((2A)/(sin hatA*sin hatB*sin hatC))$

$= \sqrt{\frac{10}{sin (\frac{1}{12} π) \cdot sin (\frac{3}{4} π) \cdot sin (\frac{1}{6} π)}}$ $=sqrt(10/(sin(1/12pi)*sin(3/4pi)*sin(1/6pi)))$

$= 10.45$ $=10.45$

Therefore,

$a = 10.45 sin (\frac{1}{12} π) = 2.71$ $a=10.45sin(1/12pi)=2.71$

$b = 10.45 sin (\frac{3}{4} π) = 7.39$ $b=10.45sin(3/4pi)=7.39$

$c = 10.45 sin (\frac{1}{6} π) = 5.23$ $c=10.45sin(1/6pi)=5.23$

The radius of the incircle is $= r$ $=r$

$\frac{1}{2} \cdot r \cdot (a + b + c) = A$ $1/2*r*(a+b+c)=A$

$r = \frac{2 A}{a + b + c}$ $r=(2A)/(a+b+c)$

$= \frac{10}{15.33} = 0.65$ $=10/(15.33)=0.65$

The area of the incircle is

$a r e a = π \cdot r^{2} = π \cdot {0.65}^{2} = 1.34 u^{2}$ $area=pi*r^2=pi*0.65^2=1.34u^2$

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A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{12}$ $pi/12$ , vertex B has an angle of $\frac{3 π}{4}$ $(3pi)/4$ , and the triangle's area is $5$ $5$ . What is the area of the triangle's incircle?

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