Can you please help me on this, I have just seen that somebody else has asked this question previously but I don't understand how you manage to get two lengths.

Trigonometric Question:-
In triangle ABC, the angle A is 42 degrees, the side AB is 16cm and the side BC is 14cm. Find the two possible values of the side AC. How do you get two lengths?

One way to solve this is to use the law of cosines, which states:


A^2 + B^2 -2AB cos(C) = C^2


Here you can substitute 16 for length A, 14 for length C and 42 degrees for angle C, and solve for length B. Because it's a quadratic equation, you'll get two values for length B. See the attached figure and you'll see how that works - the third side of the triangle can be either BC or BC'.