Question

PLEASE PLEASE HELP! Find the measure of APB^.

Answer 1

Answer:

65°

Step-by-step explanation:

Radii CA and CB are perpendicular to tangent lines AT and BT, so

$\angle CAT=\angle CBT=90^{\circ}$

Since angle BAT is equal to 65°, angle CAB has measure

$\angle CAB=90^{\circ}-65^{\circ}=25^{\circ}.$

Consider triangle ACB. This triangle is isosceles, because CA=CB as radii of the circle. Two angles adjacent to the base are congruent, thus

$\angle CBA=\angle CAB=25^{\circ}$

The sum of the measures of all interior angles in triangle is always 180°, so

$\angle CAB+\angle CBA+\angle ACB=180^{\circ}\\ \\25^{\circ}+25^{\circ}+\angle ACB=180^{\circ}\\ \\\angle ACB=130^{\circ}$

Angle ACB is central angle subtended on the minor arc AB, angle APB is inscribed angle subtended on the same minor arc AB. The measure of inscribed angle is half the measure of central angle subtended on the same arc, so

$x=\angle APB=\dfrac{1}{2}\cdot 130^{\circ}=65^{\circ}$