**A triangle has an area of 16 in^2, and two of the sides of the triangle have lengths 5 in. and 7 in. Find the angle included by these two sides?**

**Answer:**

Here’s how I would do it. Draw the triangle with the 7 as the base and draw an altitude (height). Area = (.5)base x alt = (.5)7a = 16. This gives you a = 32/7

Then looking at right triangle with 5 as hypotenuse, use sin = opp/hyp = a/5 =32/7/5 = 0.914

Then take inverse sine to get a measure of 66.1 degrees.