Q5 of 22 Page 126

O is the centre of circle in the picture beside, if ∠APB = 80°, let us find the sum of the measures of ∠AOB and ∠COD and answer with reason.

In ΔAPB, sum of all angle is equal to 180°


⇒ ∠PAB + ∠ABP = 180 - ∠APB


⇒ ∠PAB + ∠ABP = 180 – 80 = 100 --------- (1)


Also by the theorem:-


The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.


and


Substituting above values in eq. (1)



⇒ ∠AOD + ∠BOC = 200


Also, as the sum of angles around the point is equal to 360°


⇒ ∠AOD + ∠BOC + ∠AOB + ∠COD = 360


⇒ ∠AOB + ∠COD = 360 – 200 = 160°


More from this chapter

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4

In the adjacent figure O is the centre of the circle; ∠AOD = 40° and ∠ACB = 35°; let us write by calculating the value of ∠BCO and ∠BOD, and answer with reason.

 6

Like the adjoining figure, we draw two circles with centres C and D which intersect each other at the points A and B. We draw a straight line through the point A which intersects the circle with centre C at the point P and the circle with centre D at the point Q.

Let us prove that (i) ∠PBQ = ∠CAD (ii) ∠BPC = ∠BQD


 7

If the circumcentre of triangle ABC is O; let us prove that ∠OBC + ∠BAC = 90°.

 8

Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D, let us prove that ΔABCD is an equilateral triangle.