Q4 of 22 Page 126

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.

As ∠ACB = 35° and 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.



As ∠BOD = ∠AOD + ∠AOB


⇒ ∠BOD = 40 + 70 = 110°


Also since DOC forms a straight line,


⇒ ∠BOC = 180 - ∠BOD = 180 – 110 = 70°


In ΔBOC, OB = OC and as we know that angle opposite to equal sides are equal.


⇒ ∠OCB = ∠OBC


Sum of angles in a triangle is equal to 180.


⇒ 2∠BCO = 180 – 70


⇒ ∠BCO = 55°


More from this chapter

All 22 →
5

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.

 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.