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# Angles in a circle | PRMO-2018 | Problem 8

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

Answer:$80$

PRMO-2018, Problem 8

Pre College Mathematics

## Try with Hints

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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