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Question

In Figure 2, AB and AC are tangents to the circle with centre O such that $$\angle BAC = 40^{\circ}$$. Then $$\angle BOC $$ is equal to

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A
40
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B
50
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C
140
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D
150
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Solution

The correct option is C $$140^{\circ}$$
Tangent is perpendicular to radius at point of contact.
So, $$\angle ABO = \angle ACO = 90^{\circ}$$

In a quadrilateral, the sum of the angles is $$360^{\circ}$$​. 
$$\angle{BAC} + \angle{BOC} + \angle ABO + \angle ACO = 360^{\circ}$$

$$\therefore \angle{BAC} + \angle{BOC} = 18​0^{\circ}$$
                      $$\angle{BOC} = 180^{\circ} -40^{\circ} $$
                      $$\angle{BOC} = 140^{\circ}$$
So, option C is the answer

Maths

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