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 $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $28$ $28$ . What is the area of the triangle's incircle?

Question

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 $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $28$ $28$ . What is the area of the triangle's incircle?

1 Answer

Impact of this question

1397 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-01T06:23:42

The area of the incircle is $= 11.78 u^{2}$ $=11.78u^2$

Explanation:

enter image source here

The area of the triangle is $A = 28$ $A=28$

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

The angle $ˆ B = \frac{5}{12} π$ $hatB=5/12pi$

The angle $ˆ C = π - (\frac{1}{8} π + \frac{5}{12} π) = \frac{11}{24} π$ $hatC=pi-(1/8pi+5/12pi)=11/24pi$

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,

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

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

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

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

$= \sqrt{\frac{56}{sin (\frac{1}{8} π) \cdot sin (\frac{5}{12} π) \cdot sin (\frac{11}{24} π)}}$ $=sqrt(56/(sin(1/8pi)*sin(5/12pi)*sin(11/24pi)))$

$= 12.36$ $=12.36$

Therefore,

$a = 12.36 sin (\frac{1}{8} π) = 4.73$ $a=12.36sin(1/8pi)=4.73$

$b = 12.36 sin (\frac{5}{12} π) = 11.94$ $b=12.36sin(5/12pi)=11.94$

$c = 12.36 sin (\frac{11}{24} π) = 12.24$ $c=12.36sin(11/24pi)=12.24$

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

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

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

$= \frac{56}{28.91} = 1.94$ $=56/(28.91)=1.94$

The area of the incircle is

$a r e a = π \cdot r^{2} = π \cdot {1.94}^{2} = 11.78 u^{2}$ $area=pi*r^2=pi*1.94^2=11.78u^2$

Science

Math

Humanities

... and beyond

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 $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $28$ $28$ . What is the area of the triangle's incircle?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

A triangle has vertices A, B, and C. Vertex A has an angle of π8pi/8 , vertex B has an angle of 5π12(5pi)/12 , and the triangle's area is 2828 . What is the area of the triangle's incircle?

1 Answer

Explanation:

Related questions

Impact of this question

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 $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $28$ $28$ . What is the area of the triangle's incircle?