Question & Answer

In the figure XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B prove that $\angle {\text{AOB = 9}}{{\text{0}}^\circ}$.

ANSWER Verified Verified
Hint: Tangent at any point of a circle is perpendicular to the radius through the point of contact. We will join O to C and then use a congruence and linear pair of angles to prove that $\angle {\text{AOB = 9}}{{\text{0}}^\circ }$.

Complete step-by-step answer:
Now, we will join O to C and draw the figure.

Now, from the figure, we can see that OC, OP and OQ are the radiuses of the circle. Now, as the tangent is perpendicular to radius at point of contact, so $\angle {\text{OPA = 9}}{{\text{0}}^\circ }$. Also, $\angle OQC{\text{ }} = {\text{ }}\angle OCA{\text{ }} = {\text{ }}\angle OCB{\text{ }} = {\text{ }}{90^\circ }$. Also, the lengths of tangents drawn from an external point to a circle are equal. Therefore, for an external point A, PA = PC, for point B, QB = BC.
So, in triangle OQA and triangle OCA, we have
PA = PC (lengths of tangent)
OP = OC (radii)
\[\angle {\text{OPA = }}\angle {\text{OCA}}\] (each equal to ${90^\circ }$)
So, $\Delta {\text{OQA }} \cong {\text{ }}\Delta {\text{OCA}}$ by SAS property.
So, by c.p.c.t we get, $\angle {\text{POA = }}\angle {\text{COA}}$ …. (1)
Now, in triangle OQB and OCB, we have
BQ = BC (lengths of tangent)
OQ = OC (radii)
$\angle {\text{OQB = }}\angle {\text{OCB}}$ (each equal to ${90^\circ }$)
So, $\Delta {\text{OQB }} \cong {\text{ }}\Delta {\text{OCB}}$ by SAS property.
So, by c.p.c.t. we get, $\angle {\text{QOB = }}\angle {\text{COB}}$ … (2)
Now, by linear pair of angles which states that the sum of all angles in a straight line is equal to ${180^\circ }$.
So, on straight line POQ, we have
$\angle POA{\text{ }} + {\text{ }}\angle AOC{\text{ }} + {\text{ }}\angle QOC{\text{ }} + {\text{ }}\angle BOC{\text{ }} = {\text{ }}{180^\circ }$
 From equation (1) and (2), we get
$2\angle AOC{\text{ }} + {\text{ }}2\angle BOC{\text{ }} = {\text{ }}{180^\circ }$
$2\angle {\text{AOB = 18}}{{\text{0}}^\circ }$
Therefore, $\angle {\text{AOB = 9}}{{\text{0}}^\circ }$. Hence proved.

Note: When we come up with such types of problems, we have to follow a few steps. First, we will do some constructions in the figure given in the question. Then we will apply properties of tangent to proceed towards the solution. Then, we will use the congruence of triangles to prove the given statement. While solving such questions, carefully apply the properties, because most students commit mistakes while applying property.