91Ó°ÊÓ

The figure is given.

Following properties are used here,

1. In a circle, any line from center of circle to the point of contact of tangent is always perpendicular to tangent.
2. The line passing from the center and intersecting point of two tangents to its circle, always bisect the two angles, forming on the center and intersecting point.
3. In a right triangle, remaining two angles, except right angle, are complementary angles.
   Also, if the sum of two angles is $90°$, so these are complementary and if their sum is $180°$, so these angles are supplementary.

So, in given figure as $OAâŠ\yen AP,$ so $∠PAO=90°$and thus $â–\mu APO$is a right triangle.

Similarly as $OBâŠ\yen BP,$ so $∠PBO=90°.$ Hence, $â–\mu BPO$ is a right triangle. Thus, using first property, $∠PAO=90°.$ So, these two angles are congruent.

Also, the sum of these two angles will be $180°,$ so these two angles are supplementary also.

Further, in right triangle $â–\mu APO,$ as angle A is $90°,$ so by angle sum property of triangle,

$∠POA+∠APO=90°,$ Hence these two angles are complementary to each other. Similarly in right triangle $â–\mu BPO,∠POB+∠BPO=90°.$ So, these two are also complementary angles. Also,

Using third property, another sets of congruent angles are $∠POB=∠POA$and $∠APO=∠BPO.$

Therefore, it is concluded that pairs of congruent angle are

$\left(i\right)∠APO,∠BPO\left(ii\right)∠POB,∠POA\left(iii\right)∠PBO,∠PAO.$

Pairs of complementary angles are

$\left(i\right)∠APO,∠POA\left(ii\right)∠POB,∠BPO.$

Further only pair of supplementary angles:

$∠PBO,∠PAO.$