Three circles with centers A, B and C have respective radii 50, 30 and 20 inches and are tangent to each other externally. Find the area in (in^2) of the curvilinear triangle formed by the three circles

Refer to the figure below which illustrates the given problem. The yellow

shaded area is the curvilinear triangle.

First, we should find angles A, B, and C of triangle ABC.

a = 50 in, b = 70 in, c = 80 in.

From the Law of Cosines: c² = a² + b² - 2ab cosC

cos C = (a² + b² - c²) / (2ab) = (50²+70²-80²) / (2*50*70) = 0.1429

C = arcos (0.1429) = 81.79°.

From the Law of Sines,

sin (B) / b = sin (C) / c

sin (B) = (b/c) sin (C) = (70/80) * sin (81.79°) = 0.866

B = arcsin (0.866) = 60°

Because the sum of angles in a triangle is 180°, therefore

A = 180 - (60 + 81.79) = 38.21°.

Calculate the area of ΔABC from the Law of Sines.

Area = (1/2) ab sinC = (1/2) * 50*70*sin (81.79°) = 1732.065 in²

Calculate the areas of arc segments created by circle centers A, B, and C.

Aa = (38.2/360) * π (50²) = 833.39 in²

Ab = (60/360) * π (30²) = 471.2 in²

Ac = (81.79/360) * π (20²) = 285.5 in²

The area of the curvilinear triangle is

1732.065 - (833.39 + 471.2 + 285.5) = 141.975 in²

Answer: 142.0 in² (nearest tenth)