Question

Can y'all please help me I need it urgently (class 9th)
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Answer 1

Answer:

69. OA=OB (RADII OF THE SAME Circle)

Angle OBA=40degree

In Triangle OAB by ( angle sum property)

AngleAOB=180-80=100

We know angle 2ACB=AngleAOB

ACB =100÷2 =50 degree

70. In △ABD, ∠ABD + ∠ADB + ∠DAB = 180° ( angle sum property)

50° + ∠ADB + 60° = 180°

⇒ ∠ADB = 70°

Now, ∠ACB = ∠ABD = 70° [∵ angle in same segment]

71. In ΔAOB,

∠OBA = ∠BAO

[angles opposite to equal sides are equal]

∠OBA = 60° [∴ ∠BAO = 60°, given]

∠ABC=∠ADC

[angles in the same segment AC are equal]

∠ADC = 60°

72. In ∆OAB, ∠OAB + ∠ABO + ∠BOA = 180°

∠OAB + ∠OAB + 90° = 180° ⇒ 2∠OAB = 180°- 90°

[angles opposite to equal sides are equal] [angle sum property of a triangle]

⇒ ∠OAB = 90°/2 = 45° …(i)

In ΔACB, ∠ACB + ∠CBA + ∠CAB = 180°

∴ 45°+ 30°+ ∠CAB = 180°

⇒ ∠CAB = 180° – 75° = 105°

∠CAO+ ∠OAB = 105°

∠CAO + 45° = 105°

∠CAO = 105° – 45° = 60°