Maths problem help please?

Take π = 3.14

Let O₁ be the circle of radius 5 cm, and O₂ be circle of radius 12 cm.
Let AB be the common chord of the two circles.

In ΔO₁O₂A :
O₁A² + O₂A² = (5² + 12²) cm²
O₁A² + O₂A² = 169 cm²
O₁A² + O₂A² = (13 cm)²
O₁A² + O₂A² = O₁O₂²
By the inverse theorem of Pythagorean theorem, ∠O₁AO₂ = 90⁰

tan∠AO₁O₂ = 12/5
∠AO₁O₂ = 67.38°
∠AO₁B = 67.38° × 2 = 134.76°
∠AO₂B = (90° - 67.38°) × 2 = 45.24°

Area common to the two circles
= (Area of sector O₁AB) + (Area of sector O₂AB) - (Area of ΔO₁AO₂) - (Area O₁BO₂)
= [π × 5² × (134.76/360)] + [π × 12² × (45.24/360)] - [(1/2) × 5 × 12] × 2 cm²
= [3.14 × 5² × (134.76/360)] + [3.14 × 12² × (45.24/360)] - [5 × 12] cm²
= 26.2 cm² (to 2 sig. fig.)

https://en.wikipedia.org/wiki/Circular_segment

So our first step is to figure out the angles at each circle's center that the arc subtends.
Luckily, a 5-12-13 triangle is a right triangle, so we can use some easy trig to figure it out.

sin(A) / 5 = sin(90) / 13
sin(A) = 5 * 1 / 13
sin(A) = 5/13
A = arcsin(5/13)

We want 2A as our angle

sin(B) = 12 / 13
B = arcsin(12/13)

The area of a segment is: (R^2 / 2) * (t - sin(t))
R = Radius
t = angle

Add the areas together

(12^2 / 2) * (2 * arcsin(5/13) - sin(2 * arcsin(5/13))) + (5^2 / 2) * (2 * arcsin(12/13) - sin(2 * arcsin(12/13)))
144 * (arcsin(5/13) - sin(arcsin(5/13))cos(arcsin(5/13))) + 25 * (arcsin(12/13) - sin(arcsin(12/13))cos(arcsin(12/13)))

144 * arcsin(5/13) - 144 * (5/13) * sqrt(1 - (5/13)^2) + 25 * arcsin(12/13) - 25 * (12/13) * sqrt(1 - (12/13)^2)
144 * arcsin(5/13) + 25 * arcsin(12/13) - 144 * (5/13) * (12/13) - 25 * (12/13) * (5/13)
144 * arcsin(5/13) + 25 * arcsin(12/13) - (5/13) * (12/13) * (144 + 25)
144 * arcsin(5/13) + 25 * arcsin(12/13) - 60