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Try this beautiful problem from PRMO, 2018 based on Angles in a circle.

Let AB be a chord of circle with centre O. Let C be a point on the circle such that \(\angle ABC\) = \(30^{\circ}\) and

O lies inside the triangle ABC. Let D be a point on AB such that \(\angle DCO\) = \(\angle OCB\) = \(20^{\circ}\). Find the

measure of \(\angle CDO\) in degrees.

- \(75^{\circ}\)
- \(80^{\circ}\)
- \(60^{\circ}\)

Geometry

Circle

Angle

But try the problem first...

Answer:$80$

Source

Suggested Reading

PRMO-2018, Problem 8

Pre College Mathematics

First hint

We have to find out the \(\angle CDO\).If we can find out the value of \(\angle ODC\) & \(\angle COD\)...then we can easily find out the value of \(\angle CDO\)

Can you now finish the problem ..........

Second Hint

For \(\angle ODC\) & \(\angle COD\) ,we have to find all the angles of the triangles using cyclic property and given data.such as OB=OC,so \(\angle OBC=\angle OCB\).\(\angle ABC=30\) So the \(\angle ABO=30-20\)

Can you finish the problem........

Final Step

Given \(\angle OCB = 20^{\circ}\)

\(\angle OBC = 20^{\circ}\)[as OB=OC ,radius of the circle]

\(\angle OBA =\angle ABC -\angle OBC=30^{\circ}-20^{\circ}=10^{\circ}\)

\(\angle OAB = 10^{\circ}\)[as OB=OA,radius of the circle]

\(\angle BOC =180-\angle OBC-\angle OCB=180-20-20=140\),

Now \(\angle BOC =140^{\circ} \Rightarrow \angle A = 70°\)[since an arc subtends double angle compare to circumference]

\(\angle OAC=\angle BAC -\angle BAO=70-10= 60^{\circ}\)

\(\angle ACD = 40^{\circ}\)

Now C is circumcenter of \(\triangle AOD\)

as \(\angle OCD = 2\angle OAD\)

\(\angle AOD =\frac{1}{2}\angle OAD = 20^{\circ}\)

\(\angle DOC = \angle AOD + \angle AOC\)

= 20 + 60

= 80

\(\angle ODC = 180 – (\angle DOC + \angle OCD)\)

= 180 – (80 + 20)

= 80°

- https://www.youtube.com/watch?v=i_pmSwUO4LA
- https://www.cheenta.com/combination-of-cups-problem-prmo-2018-problem-11/

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