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 $\frac{3 π}{8}$ $(3pi)/8$ , and the triangle's area is $6$ $6$ . 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 $\frac{3 π}{8}$ $(3pi)/8$ , and the triangle's area is $6$ $6$ . What is the area of the triangle's incircle?

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Answer 1 · 2017-06-28T08:18:46

The area of the incircle is $= 1.89 u^{2}$ $=1.89u^2$

Explanation:

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The area of the triangle is $A = 6$ $A=6$

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

The angle $ˆ B = \frac{3}{8} π$ $hatB=3/8pi$

The angle $ˆ C = π - (\frac{2}{24} π + \frac{9}{24} π) = \frac{13}{24} π$ $hatC=pi-(2/24pi+9/24pi)=13/24pi$

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,

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

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

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

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

$= \sqrt{\frac{12}{sin (\frac{π}{12}) \cdot sin (\frac{3}{8} π) \cdot sin (\frac{13}{24} π)}}$ $=sqrt(12/(sin(pi/12)*sin(3/8pi)*sin(13/24pi)))$

$= 7.11$ $=7.11$

Therefore,

$a = 7.11 sin (\frac{1}{12} π) = 1.84$ $a=7.11sin(1/12pi)=1.84$

$b = 7.11 sin (\frac{3}{8} π) = 6.57$ $b=7.11sin(3/8pi)=6.57$

$c = 7.11 sin (\frac{13}{24} π) = 7.05$ $c=7.11sin(13/24pi)=7.05$

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

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

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

$= \frac{12}{15.5} = 0.78$ $=12/(15.5)=0.78$

The area of the incircle is

$a r e a = π \cdot r^{2} = π \cdot {0.78}^{2} = 1.89 u^{2}$ $area=pi*r^2=pi*0.78^2=1.89u^2$

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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 $\frac{3 π}{8}$ $(3pi)/8$ , and the triangle's area is $6$ $6$ . 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 $\frac{3 π}{8}$ $(3pi)/8$ , and the triangle's area is $6$ $6$ . What is the area of the triangle's incircle?