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: $8\sqrt{3}$

Question

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: $8\sqrt{3}$

Team Careers360 · Accepted Answer

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 $ riangle ABO$ and $ riangle ACO$,
$OB=OC$ (Both are radius)
$\angle B=\angle C$ (both are $90^{\circ}$)
$AB=AB$ (tangents are equal)
$ riangle ABO \cong riangle ACO$, (by SAS )
In $ riangle ACO$,
$ an 30^{\circ}=\frac{OC}{AC}$
⇒ $\frac{1}{\sqrt3}=\frac{6}{AC}$
⇒ $AC=6\sqrt3$
⇒ $AB = AC = 6\sqrt3$ cm
⇒ $AB = AC$
$ herefore\sqrt{\left ( AB^{2}+AC^{2} ight )}=AB\sqrt2=6\sqrt6$
Hence, the correct answer is $6\sqrt6$.

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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 (AB2+AC2)? Option 1: 66Option 2: 46Option 3: 93Option 4: 83

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Question : $A B$ and $A C$ are the two tangents to a circle whose radius is 6 cm. If $∠ B A C = 60^{\circ}$ , then what is the value (in cm) of $\sqrt{(A B^{2} + A C^{2})} ?$
Option 1: $6 \sqrt{6}$
Option 2: $4 \sqrt{6}$
Option 3: $9 \sqrt{3}$
Option 4: $8 \sqrt{3}$