Discussion

Question:

AB and AC are the two tangents to a circle whose radius is 6 cm. If ∠BAC = 60°, then what is the value (in cm) of $\sqrt{A B^{2} + A C^{2}}$ ?

Explanation:

OB = 6

∆ ABO is a 30°-60°-90° triangle

∴ AB = OB√3 = 6√3 cm

∴ $\sqrt{A B^{2} + A C^{2}}$ = $\sqrt{{(6 \sqrt{3})}^{2} + {(6 \sqrt{3})}^{2}}$ = 6√6 cm

Hence, option (a).

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