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)
