Question

The figure below shows a triangle with vertices A and B on a circle and vertex C outside it. Side AC is tangent to the circle. Side BC is a secant intersecting the circle at point X:
What is the measure of angle ACB?

29°

8°

16°

21°

Answer 1

Answer:

8°

Step-by-step explanation:

Let o be the center of the circle. Consider triangle AOB. This triangle is isosceles triangle, because OB = OA as radii of the circle. Angles adjacent to the base AB of the triangle AOB are congruent (as angles adjacent to the base of isosceles triangle). From the diagram, the measure of the arc AB is 100°, then the measure of the central angle AOB is 100° too. The sum of the measures of all interior angles of the triangle is always 180°, so

$m\angle OBA+m\angle OAB+m\angle AOB=180^{\circ}\\ \\m\angle OBA=m\angle OAB=\dfrac{1}{2}(180^{\circ}-100^{\circ})=40^{\circ}$

AC is tangent to the circle, this means AC is perpendicular to the radius OA, so

$m\angle OAC=90^{\circ}\\ \\m\angle BAC=m\angle OAB+m\angle OAC=40^{\circ}+90^{\circ}=130^{\circ}$

In triangle ABC,

$m\angle ACB+m\angle CBA+m\angle BAC=180^{\circ}\\ \\m\angle ACB=180^{\circ}-42^{\circ}-130^{\circ}=8^{\circ}$