TA and TB are tangents and 0 is the centre.Angle CAD =x,find in terms of x(i)angle OBA(ii)angle CPB.
To find the measures of angle OBA and angle CPB in terms of angle CAD (x), we can use the properties of tangent lines and the angles formed by tangents and the radius.
(i) Angle OBA:
In a circle, a radius that is perpendicular to a tangent line forms a right angle (90 degrees). Since TA is a tangent line, angle TAB is a right angle.
Similarly, since TB is also a tangent line, angle TBA is also a right angle.
Angle OBA can be found by subtracting angle TAB from angle TBA:
Angle OBA = angle TBA - angle TAB
Since angle TBA and angle TAB are both right angles, their sum is 180 degrees. So, we have:
Angle OBA = 180 degrees - angle TAB
Since angle TAB is x degrees (given in the question as angle CAD), we have:
Angle OBA = 180 degrees - x
Therefore, angle OBA is equal to 180 degrees minus angle CAD.
(ii) Angle CPB:
In this case, we have two tangent lines (TA and TB) intersecting outside the circle at point P. The angle formed between the tangents at the point of intersection (P) can be found using the property that "the angle between two tangents drawn from an external point is equal to half the difference of the measures of the intercepted arcs."
In this case, the intercepted arcs are arc TA and arc TB.
The measure of arc TA is equal to twice the measure of angle TAB, which is 2x degrees.
The measure of arc TB is equal to twice the measure of angle TBA, which is 2x degrees.
So, the difference between the measures of the intercepted arcs is:
arc TA - arc TB = 2x - 2x = 0 degrees
According to the tangent property, the angle between the tangents at the point of intersection is equal to half the difference of the intercepted arcs. Since the difference is 0 degrees, we can conclude that the angle CPB is also 0 degrees.
Therefore, angle CPB is equal to 0 degrees.