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In fig. AB and AC are tangents to the circle with centre O such that $\angle BAC=40°$.then $\angle BOC$ is equal to:
$( A) \ 40^{o}$
$( B) \ 50^{o}$
$( C) \ 140^{o}$
$( D) \ 160^{o}$"
Given: Here given fig.2, a circle with centre O, AB and AC are tangent to the circle on B and C. $\angle BAC=40^{o}$
To do: To find out $\angle BOC=?$
Solution:
As shown in the given fig. AB and AC are tangents to the circle with centre O.
$\therefore \ OB\bot AB$ and $OC\bot AC$
$\Rightarrow \angle OBA=90^{o}$ and $\angle OCA=90^{o}$
As given $\angle BAC=40^{o}$
Here ABOC is a quadrilateral
$( \because \ sum\ of\ all\ angles\ of\ a\ quadrilateral\ is\ always\ 360^{o} )$
$\therefore \ \angle OBA+\angle BAC+\angle OCA+\angle BOC=360^{o}$
$( on\ substituting\ the\ values\ of\ \angle OBA,\angle BAC\ and\ \angle OCA)$
$\Rightarrow \angle BOC+40^{o} +90^{o} +90^{o} =360^{o}$
$\Rightarrow \angle BOC=360^{o} -220^{o}$
$\Rightarrow \angle BOC=140^{o}$
$\therefore$ Option$( C)$ is correct.
Updated on: 10-Oct-2022
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