Question

From an external point p, tangent pa and pb are drawn to a circle with centre o. If anglepab = 50, then find angle aob

Answer 1

Answer:

<PAO=50

EXTEND PA TO Q

ANGLE OAQ= 90[TANGENT RADIUS THEOREM]

ANGLE PAB+OAB+OAQ=180[LINEAR PAIR]

AOB+140=180

AOB=40

Step-by-step explanation:

Answer 2

Answer:

100

Step-by-step explanation:

There are various ways you might go around this.

One way is to use the Alternate Segment Theorem.  Let C be any point on the circle on the opposite side of AB from P.  Then by the Alternate Segment Theorem, angle ACB = angle PAB = 50.  Next, since the angle subtended at the centre of the circle is twice the angle subtended on the arc we have

angle AOB = 2 times angle ACB = 2 x 50 = 100.

Another way would be to use some triangles.  By symmetry (tangents to circles from P are equal), the triangle PAB is isosceles, so angle PBA = angle PAB = 50.

Since the sum of the angles in a triangle is 180, we then have

angle APB = 180 - angle PAB - angle PBA = 180 - 50 - 50 = 80.

Now the tangents at A and B are perpendicular to the radii there.  Also, the sum of the interior angles of a quadrilateral is 360 (it's just made up from two triangles!).  So

angle AOB = 360 - angle APB - angle PAO - angle PBO

= 360 - 80 - 90 - 90 = 100.