1.
Measure of inscribed angle is half the measure of intercepted arc. Since the inscribed angle is intercepting semicircle,
m<BAC = 1/2 * 180 = 90 degrees.
2.
Radius drawn to the point of tangency is always perpendicular to the tangent. Therefore,
m<EBC = 90
3.
come back later.
4.
Since BC = OB and triangle EBO is a right triangle, triangle EBO is 45-45-90 special right triangle. We're given that BO(radius) = 7. So,
EO = BO * sqrt(2) = 7 sqrt(2)
5.
come back later.
6.
Same as problem #1. Measure of intercepted arc is 120. So,
m<ACB = 1/2 * 120 = 60
7. Since triangle ABC is a right triangle, (remember A was 90) m<ABC must be 30 because sum of interior angles of a triangle must add up to 180.
Let's go back to 3 and 5
Triangle ABC is 30-60-90 special right triangle with AC as short leg, AB as long leg, and BC = 14 the hypotenuse. Then,
AC = 1/2 * BC = 1/2 * 14 = 7
AB = AC * sqrt(3) = 7sqrt(3)
3. AC = 7
5. AB = 7 sqrt(3)
