Q4 - mock-test-1 - pending
Q4 : Let AB be a chord of a circle with centre O. Let C be a point on the circle such that ∠ABC = 30° and O lies inside triangle ABC. Let D be a point on AB such that ∠DCO = ∠OCB = 20°. Find the measure of ∠CDO in degrees. Solution: ∠ABC = 30° ⇒ ∠AOC = 60° Since angle subtended at circumference by chord is half of that at the center. Hence, △AOC is an equilateral triangle. Since OA = OC so opposite angles are equal, AOC is already 60 so these 2 angles will also be 60. So OA = OC = AC__________[1] △OCB is isosceles because OC = OB => ∠OCB = 20° = ∠OBC So ∠ABO = 10 △OAB is isosceles because OA = OB So ∠ABO = ∠OAB = 10 ∠CAD = ∠CAO + ∠OAB = 70 ∠DCA = ∠OCA - ∠DCO = 60 - 20 = 40 In △CAD, ∠CAD = 70, ∠DCA = 40 => ∠ADC = 70 => △CAD is an isosceles triangle. => OC = CA = CD (Using [1]) ∴ In △OCD, (OC = CD) and ∠DCO = 20 ∠ODC = 80° [Because OCD is isosceles]