Question

Find the measure of angle A in the triangle ABC.​

Answer 1

43.531

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

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

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

`arccos((14^2-12^2-20^2)/(-2*12*20))=A`

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