Kevin has edited encyclopedias, taught history, and has an MA in Islamic law/finance. He has since founded his own financial advice firm, Newton Analytical.
Since you first started working with triangles in math class, chances are you've been exposed to the formula for the area of a triangle. However, that is only good if you know the height of the triangle. This lesson shows you how to get around that.
Area of a Triangle
Real fast, how do you find the area of a triangle? If you've been taking geometry for long, you know it's A = 1/2b * h. For many people, this equation comes to mind so quickly that you don't really have to think about it. In fact, it's often something that makes a lot of sense - draw a diagonal line down a rectangle, and you get two equally sized triangles, so you simply find the area of the rectangle and then divide by two. Easy, right?
Well, not always. Chances are those triangles that you've calculated with one half base times height almost always include either a right angle or the height marked. In those instances, it's pretty straightforward to calculate the area. But what about those pesky triangles with no such help? Are we consigned to never knowing their area? Of course not! However, it is going to take a little bit of trigonometry. Namely, we're going to have to use a different formula. In these cases, the area of a triangle is = ½ * a * b * sinC.
The lower case letters are the lengths of the sides and the upper case letter is the measure of the angle indicated. More specifically, C is the angle in between sides a and b. While it may not look as pretty as one half base times height, in this lesson we'll see it is much more useful.
Using Sine to Find the Area
Before we get concerned about the math of that formula, let's step back and look at it. Really, think about it as two parts. The one half times one side should make sense, as it is lifted straight from our usual formula. But what about that second side times the sine of the third angle? What you are essentially doing there is creating an imaginary right triangle in which we know the hypotenuse and the angle and are trying to solve for the opposite side. As such, to solve for the opposite side we just multiply both sides by the hypotenuse, which gives us the opposite side. After all, this ratio of the opposite side over the hypotenuse is all that a sine is - it changes with each angle because the ratio between sides changes. The opposite side of this imaginary right triangle is the height of our real triangle. As a result, we are still calculating one half base times height, but we're going about it in a slightly different way.
Oh, one quick note about calculators. You can use either a scientific calculator or a graphing calculator to do these problems. However, make sure that you are in degrees mode, signified by DEG often. Otherwise, you will get some really messed up answer.
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Obviously, this is a bit of a departure, but one of the beautiful things about math is that we can apply formulas and mathematical facts to real world problems. So, let's say you had to calculate the area of a wing on an airplane. The paint supplier needs to know how much paint to bring in order to complete the job, and you know that they can work out the rest once you give them the area of just one wing. The wing is swept back, with lengths of the following sides: a is 7 meters, b is 16 meters and angle C is 130 degrees. So, what is the area of the top of the wing? Let's plug the numbers into our formula. Area is equal to one half times a times b times Sine of angle C. Therefore, you have one half times 7 times 16 times the sine of 130 degrees. One half times 7 times 16 is 56, and the sine of 130 degrees is .7660. Multiply it all out and you get an area of about 42.898 meters squared.
That was pretty straightforward, so let's try something a bit more complicated. Say that you are sourcing parts for an oddly shaped solar panel. However, the engineers have lost all the measurements. One comes to you and says that they have the length of two sides, as well as the measure of the angle in between those sides. The lengths of the sides are 14 meters and 17 meters, while the angle is 61 degrees.
Once again, just plug your information into the formula one half times a times b times Sine C. That means one half times 14 times 17 times sine of 61 degrees. You crunch the numbers and come up with 119 times sine of 61 degrees, which is 0.8746. The resulting answer is 104.0797 meters squared.
In this lesson, we learned how to use sine to calculate the area of a triangle without a clearly defined height. We learned that we were substituting in an imaginary right triangle to allow us to use sine in order to find the height of the existing triangle. As such, the formula is: ½ * a * b * sinC, where a and b are the adjacent sides to angle C.
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