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Question 28
$$AB$$ and $$AC$$ are the two tangents to a circle whose radius is 6 cm. If $$\angle BAC = 60^\circ$$Â then what is the value (in cm) of $$\surd(AB^2 + AC^2)?$$
Solution
In the given figure we have angle OAB=30 and OAC=30
tan 30=r/AB
AB=r/tan 30
AB=r/(1/$$\sqrt{3})$$
AB=r$$\sqrt{3}$$
AB=6$$\sqrt{3}$$
Similarly AC=6$$\sqrt{3}$$
 $$\surd(AB^2 + AC^2)$$
=$$\surd((6\sqrt{3})^2 + ((6\sqrt{3}))^2)$$
=$$6\sqrt{6}$$
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