Question

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 the triangle ABC. Let D be a point on AB such that \DCO = \OCB = 20!.

Find the measure of \CDO in degrees.​

Answer 1

Answer:

The measure of ∠CDO is 80°.

Step-by-step explanation:

Given data: As shown AB is a chord of a circle with center O. ∠ABC = 30°

O lies inside ∆ABC. Point D is given on chord AB such that ∠DCO = ∠OCB = 20°

To find: measure of ∠CDO.

Let us consider ∆OCB

OB = OC …. [radius of the circle, ∴ ∆CDO is isosceles triangle]

∴ ∠OBC = ∠OCB = 20° [ angles opposite to equal side are equal]

∠ABC = ∠OBA + ∠OBC

Or, ∠OBA = 30° - 20° = 10° …… (i)

Now, considering ∆ABO

OA = OB ….. [ radius of circle, ∴ ∆ABO is isosceles triangle]

∴ ∠OBA = ∠OAB = 10° ….. [from (i)]

In case of ∆OCB

∠COB = 180° - (∠OBC + ∠OCB)

Or, ∠COB = 180° - (20°+ 20°)

Or, ∠COB = 140°  

Since, the angle projected by a chord at the center O of the circle is always twice the angle projected at any other point on the circle.

∴ ∠CAB = ½ * ∠COB = ½ * 140° = 70°

∴ ∠CAO = ∠CAB – ∠OAB = 70° - 10° = 60°

We know ∠AOC = 2 ∠ABC = 2 * 30° = 60° and also, ∠OCA = 60°

.

Here we get, ∆OCA as equilateral triangle, as all the angles are equal and OC = CA = OA ……. (ii)

∠DCA = ∠OCA – ∠DCO = 60° - 20° = 40°

In ∆CAD,

∠CDA = 180° - (∠DCA + ∠CAD) = 180° - (70°+ 40°) = 70°

∴ CA = CD = OC ….. [ from eq. (ii) and ∠CDA and ∠CAD are equal]

Since we get OC = CD, therefore ∆OCD is isosceles triangle and ∠COD = ∠CDO.

In ∆OCD,

∴ ∠COD + ∠CDO + ∠OCD = 180°

Or, 2 * ∠CDO + 20° = 180°

Or, 2 * ∠CDO = 160°

Or, ∠CDO = 160° / 2 = 80°