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]

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