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° = 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°
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