∠OCP = ½π rad (tangent perpendicular to radius)

∠CAB = ½π rad (angle in a semicircle)

∠PCA = ∠CBA = 0.6 rad (alternate segment theorem)

∠COA = 2∠CBA
(angle at centre = 2 x angle at circumference)
= 2 × 0.6 rad
= 1.2 rad

Radius = 10cm so diameter CB = 20cm

Area of segment AC

= area of sector COA - area of △COA

= ½r²θ - ½r² sinθ

Recall area of Sector formula is ½r²θ and area of triangle is ½ab sin c, but for the isosceles △COA, a = b = r and c = θ)

= ½r² (θ - sinθ)
= ½(10²)(1.2 - sin 1.2)
= 50(1.2 - sin 1.2)
= 60 - 50 sin 1.2


Since △CAB is right angled, apply trigo

sin 0.6 = opp/hyp = AC/CB = AC/20
AC = 20 sin 0.6

Since △CPA is also right angled
(since ∠CAP = π rad - ½π rad (angles on a straight line)), apply trigo too.

(In fact there are 3 similar right angled triangles in that diagram)

tan 0.6 = PA/AC
PA = AC tan 0.6
PA = 20 sin 0.6 tan 0.6

Area of shaded

= Area of △PAC - Area of segment AC

= ½ × PA × AC - (60 - 50 sin 1.2) cm²

= ( ½ × 20 sin 0.6 tan 0.6 × 20 sin 0.6 - 60 + 50 sin 1.2) cm²

≈ 30.225 cm²

= 30.2 cm² (3s.f)