Question

in the adjoining figure, AB and AC are tangents to the circle with centre O, angle BOC=130°, then find angle BAC​

Answer 1

Answer:Tangent is perpendicular to radius at point of contact.

So, ∠ABO=∠ACO=90  

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In a quadrilateral, the sum of the angles is 360  

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​.  

∠BAC+∠BOC+∠ABO+∠ACO=360  

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∴∠BAC+∠BOC=18​0  

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                     ∠BOC=180  

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−40  

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                     ∠BOC=140  

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Step-by-step explanation:

thiss is the correct answer

Answer 2

Answer:

now since AB & AC are tangent so <ABO=<ACO=90° & <BOC=130°(GIVEN)

ABOCA will form a quadrilateral SO sum of all the angles will be 360°

<ABO+<ACO+<BOC+<BAC=360°

90°+90°+130°+<BAC=360°

310°+<BAC=360°

<BAC=50°

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