Select Page

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

## Angles in a circle | PRMO | 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.

• $75^{\circ}$
• $80^{\circ}$
• $60^{\circ}$

### Key Concepts

Geometry

Circle

Angle

But try the problem first…

Answer:$80$

Source

PRMO-2018, Problem 8

Pre College Mathematics

## Try with Hints

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°