Angles in a circle | PRMO-2018 | Problem 8

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.

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 ..........

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........

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°

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