# A triangle has vertices A, B, and C. Vertex A has an angle of pi/3 , vertex B has an angle of (5 pi)/12 , and the triangle's area is 27 . What is the area of the triangle's incircle?

##### 1 Answer
Jul 29, 2018

color(indigo)("Area of inscribed circle " = A_i = pi r^2 ~~ 16.0643

#### Explanation:

$\text{Area of } \Delta = {A}_{t} = \left(\frac{1}{2}\right) a b \sin C = \left(\frac{1}{2}\right) b c \sin A = \left(\frac{1}{2}\right) c a \sin B$

$\text{Given } \hat{A} = \frac{\pi}{3} , \hat{B} = \frac{5 \pi}{12} , \hat{C} = \frac{\pi}{4}$, A_t = 27 $a b = \frac{2 {A}_{t}}{\sin} C = \frac{54}{\sin} \left(\frac{\pi}{4}\right) \approx 76.3675$

$b c = \frac{2 {A}_{t}}{\sin} A = \frac{54}{\sin} \left(\frac{\pi}{3}\right) = 62.3538$

$c a = \frac{2 {A}_{t}}{\sin} B = \frac{54}{\sin} \left(\frac{5 \pi}{12}\right) = 55.9049$

$a = \frac{a b c}{b c} = \frac{\sqrt{76.3675 \cdot 62.3538 \cdot 55.9049}}{62.3538} = 7.8948$

$b = \frac{a b c}{c a} = \frac{\sqrt{76.3675 \cdot 62.3538 \cdot 55.9049}}{55.9049} = 9.2291$

$c = \frac{a b c}{a b} = \frac{\sqrt{76.3675 \cdot 62.3538 \cdot 55.9049}}{76.3675} = 6.7562$

$\text{Semi-perimeter } = s = \frac{a + b + c}{2} = \frac{23.8801}{2} = 11.9401$

$\text{Radius of inscribed circle } = r = {A}_{t} / s = \frac{27}{11.9401}$

color(indigo)("Area of inscribed circle " = A_i = pi r^2 = pi * (27/ 11.9401)^2 ~~ 16.0643#