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 $\frac{π}{6}$ $pi/6$ , and the triangle's area is $9$ $9$ . What is the area of the triangle's incircle?

Question

1375 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-23T21:54:39

$A r e a \approx 2.40$ $Area ~~ 2.40$

Here is a helpful reference

Given $∠ A = \frac{π}{12}, ∠ B = \frac{π}{6}$ $angleA = pi/12, angle B = pi/6$ , and the area, $Delta = 9$

Compute $angle C$ :

$angle C = pi - angleA - angleB$

$angle C = pi - pi/12 - pi/6$

$angle C = (3pi)/4$

From the reference, the square of radius of the incircle is:

$r^2 = Delta/(cot(A/2) + cot(B/2) + cot(C/2))$

The area of the incircle is:

$Area = (piDelta)/(cot(A/2) + cot(B/2) + cot(C/2))$

Substitute the values for, $A, B, C, and Delta$

$Area = (9pi)/(cot(pi/24) + cot(pi/12) + cot((3pi)/8))$

$Area ~~ 2.40$

A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 π12pi/12 , vertex B has an angle of pi/6 π6pi/6 , and the triangle's area is 9 99 . What is the area of the triangle's incircle?