A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{π}{6}$ $( pi)/6$ , and the triangle's area is $3$ $3$ . 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 $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{π}{6}$ $( pi)/6$ , and the triangle's area is $3$ $3$ . What is the area of the triangle's incircle?

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Answer 1 · 2017-06-26T09:30:40

The area of the incircle is $= 1.02$ $=1.02$ square units.

Explanation:

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The area of the triangle is $A = 3$ $A=3$

The angle $ˆ A = \frac{1}{8} π$ $hatA=1/8pi$

The angle $ˆ B = \frac{1}{6} π$ $hatB=1/6pi$

The angle $ˆ C = π - (\frac{1}{8} π + \frac{1}{6} π) = \frac{17}{24} π$ $hatC=pi-(1/8pi+1/6pi)=17/24pi$

The sine rule is

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = k$ $a/sinA=b/sinB=c/sinC=k$

So,

$a = k sin A$ $a=ksinA$

$b = k sin B$ $b=ksinB$

$c = k sin C$ $c=ksinC$

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,

$h = c sin B$ $h=csinB$

So,

$A = \frac{1}{2} k sin A \cdot c sin B = \frac{1}{2} k sin A \cdot k sin C \cdot sin B$ $A=1/2ksinA*csinB=1/2ksinA*ksinC*sinB$

$A = \frac{1}{2} k^{2} \cdot sin A \cdot sin B \cdot sin C$ $A=1/2k^2*sinA*sinB*sinC$

$k^{2} = \frac{2 A}{sin A \cdot sin B \cdot sin C}$ $k^2=(2A)/(sinA*sinB*sinC)$

$k = \sqrt{\frac{2 A}{sin A \cdot sin B \cdot sin C}}$ $k=sqrt((2A)/(sinA*sinB*sinC))$

$= \sqrt{\frac{6}{sin (\frac{π}{8}) \cdot sin (\frac{1}{6} π) \cdot sin (\frac{17}{24} π)}}$ $=sqrt(6/(sin(pi/8)*sin(1/6pi)*sin(17/24pi)))$

$= 6.29$ $=6.29$

Therefore,

$a = 6.29 sin (\frac{1}{8} π) = 2.41$ $a=6.29sin(1/8pi)=2.41$

$b = 6.29 sin (\frac{1}{6} π) = 3.15$ $b=6.29sin(1/6pi)=3.15$

$c = 6.29 sin (\frac{17}{24} π) = 4.99$ $c=6.29sin(17/24pi)=4.99$

Observe in the figure above that joining incenter with the three vertices $A, B$ $A,B$ and $C$ $C$ results in three triangles, whose bases are $a, b$ $a,b$ and $c$ $c$ and height is $r$ $r$ , the radius of incenter and therefore the sum of the area of these triangles is area of $DeltaABC$ .

Hence we have, $1/2*r*(a+b+c)=A$

$r=(2A)/(a+b+c)$

$=6/(10.55)=0.57$

The area of the incircle is

$area=pi*r^2=pi*0.57^2=1.02$ square units.

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A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{π}{6}$ $( pi)/6$ , and the triangle's area is $3$ $3$ . 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 $\frac{π}{8}$ $pi/8$ , vertex B has an angle of $\frac{π}{6}$ $( pi)/6$ , and the triangle's area is $3$ $3$ . What is the area of the triangle's incircle?