In a triangle the length of the sides, AB is 3cm and BC is 5 cm what is the length of the side AC, if the angle formed at B is 60?

Answer 1

You'll need the law of cosines for this problem.

First, make a good sketch. Draw any triangle and label the vertices A, B, and C. Putting the given information in shows that you have been given the lengths of two sides AND the measure of the angle where those two sides meet. That's the perfect condition for the Law of Cosines.

Call the length of side AC "x" since it's the unknown value.

x² = 3² + 5² - 2(3)(5)cos60

The left side of the square of the unknown. The right side is the sum of the squares of the two known sides (3 and 5) minus the product of 2, the two known sides (3 and 5), and the "included" angle measure (60).

Now you have two choices: type the whole right side into the calculator,then take the square root to solve OR without a calculator, recognize that cos60 = 1/2. So,

x² = 3² + 5² - 2(3)(5)cos60

x² = 3² + 5² - 2(3)(5)(1/2)

x² = 9 + 25 - 15

x² = 19

x = √19

I hope that helps you. :)