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8. In Fig. 8.67, O is the centre of the circular arc
ABC. Find the angles of AABC.
300
40°
Fig. 8.67
Answers
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8. In Fig. 8.67, O is the centre of the circular arc
ABC.
Consider the attached figure while going through the following steps.
Given,
∠ COB = 30°
∠ AOB = 40°
In, Δ OCB,
∠ C = 90° (figure)
In Δ OCD,
∠ O + ∠ C + ∠ D = 180°
30° + ∠ C + 90° = 180°
120° + ∠ C = 180°
∠ C = 180° - 120°
∴ ∠ C = 60°
In Δ ABC,
∠ C = 90° - 60° = 30°
∠ A = 90° - 50° = 40°
∠ A + ∠ B + ∠ C = 180°
40° + ∠ B + 30° = 180°
70° + ∠ B = 180°
∠ B = 180° - 70°
∴ ∠ B = 110°
In, Δ OAB,
∠ A = 90° (figure)
In Δ OAD,
∠ O + ∠ A + ∠ D = 180°
40° + ∠ A + 90° = 180°
130° + ∠ A = 180°
∠ A = 180° - 130°
∴ ∠ A = 50°
Therefore, ∠ ABC = 110°
Answer:
Angle ABC=145 degree
Step-by-step explanation:
In triangle AOC,
OA=OC(radii)
Therefore, OCA=OAC
OCA+OAC+AOC=180
OCA=55= OAC
In triangle OAB,
OA=OB(radii)
Therefore,OBA=OAB
AOB+OBA+OAB=180(angle sum property)
OBA=70
OAB=AOC+CAB
CAB=15
In triangle OCB,
OC=OB
Therefore,OCB=OBC
OCB+OBC+BOC=180
OBC=75=OCB
OCB=OCA+ACB
ACB=20
ABC=ABO+CBO=145