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

A
40
B
50
C
140
D
150

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 answerMaths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More