Question

If two tangents inclined at an angle of ${60}^{\circ },$ are drawn to a circle of radius $3cm$, the length of each tangent is equal to:

• A. $3$
• B. $6$
• C. $\left(\frac{3}{2}\right)\sqrt{3}$
• D. $3\sqrt{3}$

Answer 1

The correct option is D

$3\sqrt{3}$

Explanation for the correct option.

Step $1$: Find the value of $\angle APO$ and $\angle CPO$.

Let two tangents originate from one point $P$and touch the circle at points $A$ and $C$with the center$O.$

We know that tangents through an external point to a circle are equal.

Thus, $PA=PC$

Since, the tangent to any circle is perpendicular to the radius of the circle at the point of contact.

Thus, $\angle OAP=\angle OCP=90°$

In $∆OAP$ and $∆OCP$

$OA=OC$ [ radius ]

$\angle OAP=OCP$ $\left[90°\right]$

$OP=OP$ [ Common side ]

By $RHS$ criterion,

$∆OAP$ and $∆OCP$ are congruent to each other.

Thus, $\angle APO=\angle CPO$.

Since, $\angle APC=60°$ and $\angle APC=\angle APO+\angle CPO$

$$\begin{gathered} \Rightarrow \angle APO+\angle CPO=60° \\ \Rightarrow 2\angle APO=60°\left[\angle APO=\angle CPO\right] \\ \Rightarrow \angle APO=30° \\ \Rightarrow \angle CPO=30° \end{gathered}$$

Step $2$: Compute the required length.

Given: $OA=OC=3cm$

In $∆OAP$, $\angle APO=30°$

$\frac{OA}{PA}=\tan 30°$ $\left[\tan \left(\theta \right)=\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}\right]$

$$\begin{gathered} \Rightarrow PA=\frac{OA}{\tan {30}^{\circ }} \\ =\frac{3}{{\displaystyle \frac{1}{\sqrt{3}}}} \\ =3\sqrt{3} \end{gathered}$$ $\left[\tan \left(30°\right)=\frac{1}{\sqrt{3}}\right]$

Similarly, in $∆OCP$, $\angle CPO=30°$

$\frac{OC}{PC}=\tan 30°$ $\left[\tan \left(\theta \right)=\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}\right]$

$$\begin{gathered} \Rightarrow PC=\frac{OC}{\tan {30}^{\circ }} \\ =\frac{3}{{\displaystyle \frac{1}{\sqrt{3}}}} \\ =3\sqrt{3} \end{gathered}$$ $\left[\tan \left(30°\right)=\frac{1}{\sqrt{3}}\right]$

Thus, $PA=PC=3\sqrt{3}$

Hence, option $D$ is the correct answer.