Q. 15.0( 2 Votes )

In all the pictures given below, O is the centre of the circle and A, B, C are points on it. Calculate all angles of Δ ABC and Δ OBC in each.


Answer :

Case1: The angle got by joining any point on the larger part of the circle to the end of the chord is half the angle got by joining the centre of a circle to these end.



● Join A to the centre O of the circle. This line splits the angle at A into two parts, x° and y° .


● Here Δ AOB and Δ AOC are isosceles triangle


Thus, ABO = OAB = x° and ACO = OAC = y°



● Let us take BOC = c°


(180° – 2x) + (180° – 2y) + c° = 360°


360° –2 (x+y) +c° = 360°


2(x+y) = c°



Thus,



Case2: The angle got by joining any point on the smaller part of the circle to the end of the chord is half the angle at the centre subtracted from 180°.



● Join A to the centre O of the circle. This line splits the angle at A into two parts, x° and y° .


● Here Δ AOB and Δ AOC are isosceles triangle


Thus, ABO = OAB = x° and ACO = OAC = y°



● Let us take BOC = c°


(180° – 2x) + (180° – 2y) = c°


360° –2 (x+y) = c°


2(x+y) =360° – c°



Thus,



Now, we will consider our question,


We need to find all angles of Δ ABC and Δ OBC


Consider first diagram:




Join OA, Δ AOB and Δ AOC are isosceles triangle


Thus, ABO = OAB = 20°


and ACO = OAC = 30°


Hence, BAC = OAB + OAC


BAC = 20° + 30° = 50°


From case 1, we know that,


2 BAC = BOC


BOC = 2 × 50° = 100° ….(1)


In Δ BOC


Since, OB and OC are radius of a circle


Therefore, OBC = OCB (since, angles opposite to equal side


are equal) …(2)


By, angle sum property,


OBC + OCB + BOC = 180°


OBC + OBC + 100° = 180° (from (1) and (2))


2 OBC = 180° – 100° = 80°


OBC = 40°


Thus, OBC = OCB = 40° (from (2))


Thus, ABC = ABO + OBC


ABC = 20 + 40 = 60°


Thus, ACB = ACO + OCB


ABC = 30 + 40 = 70°


Consider second diagram



In Δ AOC


Since, OA and OC are radius of a circle


Therefore, OAC = OCA = 40° (since, angles opposite to


equal side are equal) …(1)


By, angle sum property,


OAC + OCA + AOC = 180°


40° + 40° + AOC = 180° (from (1) )


AOC = 180° – 80° = 100° …(2)


From case 1, we know that,



2 ABC = AOC


(from (2) ) …(3)


Join OB,


As Δ ABO and Δ OBC are isosceles triangle


Thus, ABO = OAB = x°


and BCO = OBC = 30°


Hence, ABC = ABO + OBC


50° = + 30°


x° = 50° –30° = 20°


Thus, ABO = OAB = 20°


Thus, BAC = BAO + OAC


BAC = 20 + 40 = 60°


Thus, BCA = ACO + OCB


BCA = 30 + 40 = 70°


Consider third diagram,



Here, AOB = 40° + 70° = 110°


From case 2, we get,



Thus,


⇒∠BCA=180°– 55° = 125°


Here Δ AOC and Δ BOC are isosceles triangle


Thus, CAO = OCA = x° and BCO = OBC = y°



Also, AOC = 180 – 2x


and BOC = 180 – 2y (seen in case 2 above)


Thus, 40° = 180 – 2x and 70° = 180 – 2y


2x = 180 – 40 = 140


x = 70°


Thus, CAO = OCA = x° = 70°


And


2y = 180 – 70 = 110


y = 55°


Thus, BCO = OBC = y° = 55°


Also, In Δ OAB


AOB = 40 + 70 = 110°


Since, OA and OC are radius of a circle


Therefore, OAB = OBA (since, angles opposite to


equal side are equal) ..(1)


By, angle sum property,


OAB + OBA + AOB = 180°


OAB + OAB + 110° = 180° (from (1))


2 OAB = 180° – 110° = 70°


….(2)


Thus, OAB = OBA = 35° (from (1) and (2))


Hence, BAC = OAC – OAB


BAC = 70 – 35 = 35°


Also, ABC = OBC – OBA


BAC = 55 – 35 = 20°


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