Let O be the center of the circle. Draw BC and radius OC. As A and B are opposite ends of a diameter and C is a point on the circumference, ∠BCA = 90°. As ∠CAB and ∠COB cover the same arc CB but A is on the circumference while O is the center, ∠COB = 2•∠CAB = 2(30°) = 60°. As OC = OB = r and ∠COB = 60°, then ∆COB is an equilateral triangle, ∠BCO = ∠OBC = 60°, and BC = OC = OB = r. As OA = OC = r, ∆AOC is an isosceles triangle and ∠OCA = ∠CAO = 30°. Draw OD, where D is the point on AC where OD is perpendicular to AC. As AC is a chord and O is the center, sonce OD is perpendicular to AC, it bisects AC and AD = DC = 5/2. As ∠OCA = ∠CAO = 30°, ∠ODA = ∠CDO = 90°, and OA = OC = r (among other equalities), ∆ODA and ∆CDO are congruent 30-60-90 special right triangles. cos(30°) = DC/OC √3/2 = (5/2)/r √3r = 10/2 = 5 r = 5/√3 sin(30°) = OD/OC 1/2 = OD/(5/√3) OD = (5/√3)/2 = 5/2√3 The purple shaded area is equal to the sum of the areas of triangle ∆AOC and sector COB. Purple shaded area: Aₚ = bh/2 + (θ/360°)πr² Aₚ = 5(5/2√3)/2 + (60°/360°)π(5/√3)² Aₚ = 25/4√3 + (π/6)(25/3) Aₚ = 25√3/12 + 25π/18 Aₚ = (25/6)(√3/2+π/3) ≈ 7.97 sq units
You can find R without using sin. ABC is a 30-60-90 triangle. To get the shorter side, divide 5 by sqrt(3). The diameter ist 3 times the shorter side. The radius is half of two times the shorter side - or 5 over sqrt(3). No sin needed.
*We will avoid Pythagoras theorem at all* ! Let O is the center of the circle and OC = R (radius) < OAC=30° => 5=R√3 => *R=5/√3* and it’s abscess *a₃= R/2* Now shaded area = area Δ AOC + area (circular sector with center O and arc BC) = 1/2 λ₃⋅a₃+(πR ²⋅θ°)/360° = 1/2 R√3⋅R ²+(πR^2⋅60°)/360° = (R ² √3)/4+(πR²/6) = R²(√3/4+π/6) = (25/3)(3√3+2π)/12
Looks like it is something to do because it shows that regardless of angle there is a way to calculate the inscribed triangle and sector. Starting a simple circle theorem and describe the separated triangle and sector because of said circle theorem. I am going to test myself on ALL of the videos that have the title, "You should be able to do this" and make sure that my knowledge of geometry is impeccable.
Method using Thales theorem, trigonometric ratio, Pythagoras theorem: 1. Let O be centre of circle on diameter AB. 2. Joint BC to form ∆ABC. ∆ABC is right-angled triangle in semicircle. (Thales theorem) 3. In right-angled ∆ABC BC = AC/tan30 = 5/√3 AB^2 = AC^2 + BC^2 (Pythagoras theorem) = 5^2 + (5/√3)^2 = 100/3 4. Let R be radius of circle. R = AB/2 = 5/√3 5. Angle of sector BOC = 30 + 30 = 60 (exterior angle of isosceles ∆AOC) 6. Area of shaded region = area of ∆AOC + area of sector BOC = (1/2)(R^2)(sin120) + (π)(R^2)(60/360) = (R^2)[(√3/4) + (π/6)] = (5/√3)^2[(3√3 + 2π)/12] = 25 [(3√3 + 2π)/36]
