# A triangle has vertices A, B, and C. Vertex A has an angle of pi/8 , vertex B has an angle of ( pi)/3 , and the triangle's area is 18 . What is the area of the triangle's incircle?

$7.414 \setminus \setminus {\textrm{u n i t}}^{2}$

#### Explanation:

Given that in $\setminus \Delta A B C$, $A = \setminus \frac{\pi}{8}$, $B = \frac{\setminus \pi}{3}$

$C = \setminus \pi - A - B$

$= \setminus \pi - \setminus \frac{\pi}{8} - \frac{\setminus \pi}{3}$

$= \frac{13 \setminus \pi}{24}$

from sine in $\setminus \Delta A B C$, we have

$\setminus \frac{a}{\setminus \sin A} = \setminus \frac{b}{\setminus \sin B} = \setminus \frac{c}{\setminus \sin C}$

$\setminus \frac{a}{\setminus \sin \left(\setminus \frac{\pi}{8}\right)} = \setminus \frac{b}{\setminus \sin \left(\frac{\setminus \pi}{3}\right)} = \setminus \frac{c}{\setminus \sin \left(\frac{13 \setminus \pi}{24}\right)} = k \setminus \setminus \textrm{\le t}$

$a = k \setminus \sin \left(\setminus \frac{\pi}{8}\right) = 0.383 k$

$b = k \setminus \sin \left(\frac{\setminus \pi}{3}\right) = 0.866 k$

$c = k \setminus \sin \left(\frac{13 \setminus \pi}{24}\right) = 0.991 k$

$s = \setminus \frac{a + b + c}{2}$

$= \setminus \frac{0.383 k + 0.866 k + 0.991 k}{2} = 1.12 k$

Area of $\setminus \Delta A B C$ from Hero's formula

$\setminus \Delta = \setminus \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

$18 = \setminus \sqrt{1.12 k \left(1.12 k - 0.383 k\right) \left(1.12 k - 0.866 k\right) \left(1.12 k - 0.991 k\right)}$

$18 = 0.1644 {k}^{2}$

${k}^{2} = 109.4506$

Now, the in-radius ($r$) of $\setminus \Delta A B C$

$r = \setminus \frac{\setminus \Delta}{s}$

$r = \setminus \frac{18}{1.12 k}$

Hence, the area of inscribed circle of $\setminus \Delta A B C$

$= \setminus \pi {r}^{2}$

$= \setminus \pi {\left(\frac{18}{1.12 k}\right)}^{2}$

$= \setminus \frac{324 \setminus \pi}{1.2544 {k}^{2}}$

$= \setminus \frac{811.4445}{109.4506} \setminus \quad \left(\setminus \because {k}^{2} = 109.4506\right)$

$= 7.414 \setminus \setminus {\textrm{u n i t}}^{2}$

Jul 21, 2018

Area of Incircle color(indigo)(A_i = pi r^2 = pi * (1.5357)^2 ~~ 7.4087

#### Explanation:

$\hat{A} = \frac{\pi}{8} , \hat{B} = \frac{\pi}{3} , \hat{C} = \frac{13 \pi}{24} , {A}_{t} = 18$

${A}_{t} = \frac{1}{2} a b \sin C = \frac{1}{2} b c \sin A = \frac{1}{2} c a \sin B$

$a b = \frac{2 {A}_{t}}{\sin} C = \frac{36}{\sin} \left(\frac{13 \pi}{24}\right) \approx 36.3106$

$b c = \frac{2 {A}_{t}}{\sin} A = \frac{36}{\sin} \left(\frac{\pi}{8}\right) \approx 94.0725$

$c a = \frac{2 {A}_{t}}{\sin} B = \frac{36}{\sin} \left(\frac{\pi}{3}\right) \approx 41.5692$

$a = \frac{\sqrt{a b \cdot b c \cdot c a}}{b c} = \sqrt{\frac{a b \cdot c a}{b c}}$

$a = \sqrt{\frac{36.3106 \cdot 41.5692}{94.0725}} \approx 4$

Similarly, $b = \sqrt{\frac{36.3106 \cdot 94.0725}{41.5692}} \approx 9.0649$

$c = \sqrt{\frac{94.0725 \cdot 41.5692}{36.3106}} \approx 10.3777$

Semi perimeter $s = \frac{a + b + c}{2} = \frac{23.4426}{2} = 11.7213$

Incircle radius $r = {A}_{t} / s = \frac{18}{11.7213} \approx 1.5357$

Area of Incircle ${A}_{i} = \pi {r}^{2} = \pi \cdot {\left(1.5357\right)}^{2} \approx 7.4087$