# In all the pictur

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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