A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{3}$ $pi/3$ , vertex B has an angle of $\frac{5 π}{12}$ $(5 pi)/12$ , and the triangle's area is $16$ $16$ . 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 $\frac{π}{3}$ $pi/3$ , vertex B has an angle of $\frac{5 π}{12}$ $(5 pi)/12$ , and the triangle's area is $16$ $16$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-01-09T10:56:33

Area of I’m circle $A_{i} = 9.19 s q . U n i t s$ $color(green)(A_i = 9.19) sq. Units$

Explanation:

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Given $A = \frac{π}{3}, B = \frac{5 π}{12}, C = π - A - B = \frac{π}{4}$ $A = pi /3, B = (5pi) / 12, C = pi - A - B = pi / 4$

Area of triangle $A_{t} = 16$ $A_t = 16$

$a \cdot b = \frac{A_{t}}{(\frac{1}{2}) sin C} = \frac{16}{0.5 sin (\frac{π}{3})} \approx 36.95$ $a*b = A_t / ((1/2) sin C) = 16 / (0.5 sin (pi/3)) ~~ 36.95$

Similarly,
$b \cdot c = \frac{16}{0.5 sin A} = \frac{16}{0.5 \cdot sin (\frac{5 π}{12})} \approx 33.13$ $b*c = 16 / (0.5 sin A) = 16 / (0.5 * sin ((5pi)/12)) ~~ 33.13$

$c \cdot a = \frac{16}{0.5 sin B} = \frac{16}{0.5 \cdot sin (\frac{π}{4})} \approx 45.25$ $c*a = 16 / (0.5 sin B) = 16 / (0.5 * sin (pi/4)) ~~ 45.25$

$(a b \cdot b c \cdot c a) = {(a b c)}^{2} = 36.95 \cdot 33.13 \cdot 45.25 = 55392.95$ $(ab * bc * ca) = (abc)^2 = 36.95 * 33.13 * 45.25 = 55392.95$

$a b c = \sqrt{55392.95} \approx 23536$ $abc = sqrt(55392.95) ~~ 235 36$

$a = \frac{a b c}{b c} = \frac{235.36}{33.13} \approx 7.1$ $a = (abc) / (bc) = 235.36 / 33.13 ~~ 7.1$

$b = \frac{a b c}{c a} = \frac{235.36}{45.25} \approx 5.2$ $b = (abc) / (ca) = 235.36 / 45.25 ~~ 5.2$

$c = \frac{a b c}{a b} = \frac{235.36}{36.95} \approx 6.37$ $c = (abc) / (ab) = 235.36 / 36.95 ~~ 6.37$

Semi perimeter $\frac{S}{2} = \frac{a + b + c}{2} = \frac{7.1 + 5.2 + 6.37}{2} = 9.34$ $S/ 2 = (a + b + c) / 2 = (7.1 + 5.2 + 6.37) / 2 = 9.34$

In radius $r = \frac{A_{t}}{\frac{S}{2}} = \frac{16}{9.34} = 1.71$ $r = A_t / (S/2) = 16 / 9.34 = 1.71$

Area of in circle $A_{i} = π r^{2} = π \cdot {(1.71)}^{2} = 9.19$ $A_i = pi r^2 = pi * (1.71)^2 = color(green)(9.19)$

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A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{3}$ $pi/3$ , vertex B has an angle of $\frac{5 π}{12}$ $(5 pi)/12$ , and the triangle's area is $16$ $16$ . 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 $\frac{π}{3}$ $pi/3$ , vertex B has an angle of $\frac{5 π}{12}$ $(5 pi)/12$ , and the triangle's area is $16$ $16$ . What is the area of the triangle's incircle?