Question

Please help me solve out the subpart (iii) and (iv) of the question in the image.

Answer 1

Solution:
FOR QUESTION (III)
In ∆OAB,
OB = OA = Radius of the circle.
Therefore, we can say that ∆OAB is an isosceles triangle.
Therefore,
Angle OBA = Angle OAB = (180-140)/2 = 20°

Hence, Angle OAB = 20°

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FOR QUESTION (IV)
Consider ∆OAC,
OA = OC = Radius of the circle.
Therefore, we can say that ∆OAC is an isosceles triangle.
Angle OBA = Angle OAB = 50°
Therefore, Angle AOC = 180-50-50 = 80°

Now, in ∆OBC,
Angle BOC = 140-80 = 60°
Since, Angle OBA = 20°
Angle OBC = Angle OBA + Angle CBA .......(1)
= (20 + x)° .........(say angle CBA = x°)
Also OB = OC = Radius of the circle.
Therefore, we can say that ∆OBC is an isosceles triangle.
Therefore, we can say that,
Angle OBC = Angle OCB = (20+x)°
We know that a triangle has 180°angle,
Therefore,
Angle BOC = Angle OBC = Angle OCB = 180°
or 60° + (20+x)° + (20+x)° = 180°
or 60 + 40 + 2x = 180
or x = 80/2
or x = 40°
Therefore,
Angle OBC = Angle OCB = 20+40 = 60° ........(2)
Therefore, Angle CBA = 40° .......(From (1) & (2))

Hence, Angle CBA = 40°

Answer 2

Answer:

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