Question

The figure below shows a triangle ABC inscribed in a circle. AC = 10cm, BC = 7cm and AB = 10cm.

1. Find the size of angle BAC. 
2. Find the radius of the circle. 
3. Hence calculate the area of the shaded region.

Answer 1

1. a² =b²+ c² – 2bc cos A
   7² = 10² + 8² - (2 x 10 x 8) cos A
   49=164- 60 cos A
   -116 =160 cos A
   Cos A =¹¹⁶/₁₆₀
   Cos A = 0.725
   Cos¹ 0.725 = 43.53115
   <BAC = 43.53
    
2. ^a/_{(sin sin A )} = 2R
   ⁷/_{(sin 43.53)} = 2R
   R =5.082cm
    
3. Sin 43.53 = ^3.5/_r
   r = ^3.5/_{(sin sin 43.53 )}= 5.082 cm
   Area of △OCB =¹/₂ ab sin θ
   = ¹/₂ x 5.082 sin 87.06
   = 12.896cm²
   Area of sector ACB
   =^θ/₃₆₀ πr²
   =^87.06/₃₆₀ x ²²/₇ x 5.082 = 19.630
   Shaded region
   (19.630 — 12.896) = 6.734 cm²