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° = x° + 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°