Review question

# Given two tangents to a circle, can we find these lengths? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9792

## Solution

In the diagram, $OA$ and $OB$ are tangents to the circle, centre $C$ and radius $\quantity{4.5}{in.}$ . The lines $OA$, $OB$ are inclined at $55^\circ$ and $25^\circ$ respectively to the straight line $XOY$.

Calculate

1. the angle $COY$;

The angle $AOB$ is $100^\circ$, as together with the given angles $55^\circ$ and $25^\circ$ it adds up to $180^\circ$. We now show that the line $OC$ bisects the angle $AOB$.

Consider the two triangles $CDO$ and $COE$ here:

The lines $CD$ and $CE$ are radii of the circle. As the tangents to the circle are perpendicular to the radii $CD$ and $CE$, the triangles are right-angled at $D$ and $E$.

So the two angles are congruent, and corresponding angles and sides have the same size.

In particular, the angles $DOC$ and $EOC$ are equal. Therefore the line $OC$ is an angle bisector and, as the angle $AOB$ is $100^\circ$, angle $EOC$ is $50^\circ$.

Thus the angle $COY$ is $75^\circ$.

1. the length $OC$;

Let’s draw the above triangle $CDO$.

Now by using $\sin\alpha = \dfrac{\text{opposite}}{\text{hypothenuse}}$, we get $\sin50^\circ = \dfrac{4.5}{OC}$ and therefore $OC = \dfrac{4.5}{\sin50^\circ} \approx 5.87$. So the length $OC$ is approximately $\quantity{5.87}{in.}$ .

1. the distance of $C$ from the line $XOY$.

Let’s draw the shortest distance between $C$ and the line $XOY$.

From part (i), we have that the angle $COY$ is $75^\circ$, and from part (ii) that $OC$ is $\dfrac{4.5}{\sin50^\circ}$ (notice that we are using full accuracy here).

Calling the required distance $v$, we have

\begin{align*} \sin(75^\circ) & = \frac{v}{\left(\dfrac{4.5}{\sin50^\circ}\right)} \\ & = \frac{v\sin50^\circ}{4.5}. \end{align*}

So $v = \dfrac{4.5 \sin75^\circ}{\sin50^\circ} \approx \quantity{5.67}{in.}$