Find the measure of angle a in the triangle abc.

43.531

Step-by-step explanation:

This problem needs the law of cosines since the three sides of the triangle are involved as well as one angle.  The law of cosines is:

c^2 = a^2 + b^2 - 2ab*cos(C)

Where the lowercase letters are side values and capital letters are angle values.  Just in case I will mention side a is the one with a length of 14, side b is 20 and c = 12.

Since it is asking for angle A instead of angle C we can rewrite the law of cosines to fit that, basically just rearranging the letters.

a^2 = c^2 + b^2 - 2cb*cos(A)

Now we just plug in and solve.

14^2 = 12^2 + 20^2 - 2*12*20*cos(A)

Rearrange to get A by itself

Now we take the inverse cosine, or arccos of both sides to get our answer.

Let me know if there is something in my explanation you don't understand.  

pls refer ro the top for answer

graph b

step-by-step explanation:

one solution- graph a (clear intercepting point.)

no solution- graph b (they're parallel, there will never be a point of interception.)

infinite solutions- graph c (all points intercept.)