Mab=120 and ac is the diameter of this circle. what is the degree measure or

  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)