Question

If two tangents are inclined at an angle of ${{60}^{\circ }}$ are drawn to a circle of radius 3cm, then length of each tangent is:
A) $\dfrac{3}{2}\sqrt{3}cm$
B) $6cm$
C) $3cm$
D) $3\sqrt{3}cm$

Answer 1

Hint: The use of trigonometric ratios is required here. We can join the intersection point of the tangents with the center and make the 2 formed triangles congruent in order to get equal halves of ${{60}^{\circ }}$ in the two triangles. As the angle between a tangent and a radius is always a right angle, we can use trigonometric ratios to get our required answer.

Complete step-by-step answer:
In this solution, we are going to use the R-H-S congruence criterion which states that,
If in two right angle triangles the hypotenuses and either of the sides other than the hypotenuses are equal, then the two triangles are congruent with all corresponding parts equal.
It is given that the tangents are inclined at an angle of ${{60}^{\circ }}$. The tangents are to a circle of radius 3cm. According to this information we can construct the following figure.

Here, AB and BC are the tangents, $\angle ABC={{60}^{\circ }}$,
$\angle BAO=\angle BCO={{90}^{\circ }}$ (angle formed by a tangent and a radius of the same circle is always a right angle) and
\[AO=OC=3cm\] (radii of the same circle are equal).
Let us join B and O to form BO.
Now, in $\Delta ABO\text{ and }\Delta CBO,$
$\angle BAO=\angle BCO={{90}^{\circ }}$
\[AO=OC=3cm\]
$BO=BO\text{ }(common)$
So, $\Delta ABO\cong \Delta CBO\text{ }(by\text{ }R-H-S\text{ }congruence\text{ }criterion)$
$\Rightarrow \angle ABO=\angle CBO(CPCT).................(1.1)$
Now,
 $\begin{align}
  & \angle ABC=\angle ABO+\angle CBO={{60}^{\circ }}(Given) \\
 & \Rightarrow 2\angle ABO={{60}^{\circ }}\text{ }(From\text{ }eq(1.1)) \\
 & \Rightarrow \angle ABO={{30}^{\circ }}=\angle CBO \\
\end{align}$
$\begin{align}
  & In\text{ }\Delta ABO, \\
 & \tan (\angle ABO)=\dfrac{AO}{AB} \\
 & \Rightarrow \tan ({{30}^{\circ }})=\dfrac{3}{AB} \\
 & \Rightarrow AB=\dfrac{3}{\tan ({{30}^{\circ }})} \\
 & \Rightarrow AB=\dfrac{3}{\dfrac{1}{\sqrt{3}}} \\
 & \Rightarrow AB=3\sqrt{3}cm \\
\end{align}$
Now, $AB=3\sqrt{3}=BC$ (tangents from a single external point to the same circle are equal)
Therefore, we obtain our required answer as option (d)$3\sqrt{3}$.

Note: In place of tangent we could also have used cotangent and would have arrived at the same answer as we got by using tangent for the calculation of our needed answer.