Question : $AB$ and $AC$ are the two tangents to a circle whose radius is 6 cm. If $\angle BAC=60^{\circ}$, then what is the value (in cm) of $\sqrt{\left ( AB^{2}+AC^{2} \right )}?$
Option 1: $6\sqrt{6}$
Option 2: $4\sqrt{6}$
Option 3: $9\sqrt{3}$
Option 4: $8\sqrt{3}$
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Correct Answer: $6\sqrt{6}$
Solution :
In the given problem, $AB$ and $AC$ are tangents to a circle from an external point $A$, and the angle between them is $60^{\circ}$. The radius of the circle is 6 cm. The triangle is formed by the two tangents and the line joining the centre of the circle and the external point is right-angled. Such that $AB = AC$. In $\triangle ABO$ and $\triangle ACO$, $OB=OC$ (Both are radius) $\angle B=\angle C$ (both are $90^{\circ}$) $AB=AB$ (tangents are equal) $\triangle ABO \cong\triangle ACO$, (by SAS ) In $\triangle ACO$, $\tan 30^{\circ}=\frac{OC}{AC}$ ⇒ $\frac{1}{\sqrt3}=\frac{6}{AC}$ ⇒ $AC=6\sqrt3$ ⇒ $AB = AC = 6\sqrt3$ cm ⇒ $AB = AC$ $\therefore\sqrt{\left ( AB^{2}+AC^{2} \right )}=AB\sqrt2=6\sqrt6$ Hence, the correct answer is $6\sqrt6$.
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