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The Law of Sines

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**What if the triangle we want to solve is NOT a right triangle**

What if the triangle we want to solve is NOT a right triangle? In the next two sections we’ll develop ways of solving these triangles if we know at least one side and two other pieces of info (sides or angles or one of each). c a b We’ll label these triangles with sides a, b and c and use their Greek alphabet counterparts for the angles opposite those sides as shown above.

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**Draw a perpendicular line and call the length h**

Draw a perpendicular line and call the length h. We do this so that we have a right triangle which we already know how to work with. c a h b Let’s write some trig functions we know from the right triangles formed. Solve these for h Since these both = h we can substitute divide both sides by ac ac ac

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**THE LAW OF SINES THE LAW OF SINES**

This process can be repeated dropping a perpendicular from a different vertex of the triangle. What we get when we combine these is: THE LAW OF SINES THE LAW OF SINES Use these to find missing angles Use these to find missing sides What this says is that you can set up the ratio of the sine of any angle in a triangle and the side opposite it and it will equal the ratio of the sine of any other angle and the side opposite it. If you know three of these pieces of information, you can then solve for the fourth.

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There are three possible configurations that will enable us to use the Law of Sines. They are shown below. You don’t have an angle and side opposite it here but can easily find the angle opposite the side you know since the sum of the angles in a triangle must be 180°. ASA SAA You may have an angle, a side and then another angle You may have a side and then an angle and then another angle What this means is that you need to already know an angle and a side opposite it (and one other side or angle) to use the Law of Sines. SSA You may have two sides and then an angle not between them.

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**Solve a triangle where = 55°, = 82° and c = 9**

Draw a picture (just draw and label a triangle. Don't worry about having lengths and angles look right size) 9 c 55 a 6.20 This is SAA 43 82 Do we know an angle and side opposite it? If so the Law of Sines will help us determine the other sides. 7.44 b How can you find ? Hint: The sum of all the angles in a triangle is 180°. How can you find a? (Remember it is NOT a right triangle so Pythagorean theorem will not work). You can use the Law of Sines again.

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**Let's look at a triangle where you have SSA.**

b It could be that you can't get the sides to join with the given info so these would be "no solution". a It is easy to remember that the tricky ones are SSA (it's bassakwards). b It could be that there is one triangle that could be formed and you could solve the triangle. a b b a a a, b and are the same in both of these triangles. It could be that since side c and are not given that there are two ways to draw the triangle and therefore 2 different solutions to the triangle

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**Solve a triangle where = 95°, b = 4 and c = 5**

You can just check to see if there are two triangles whenever you have the SSA case. The "no solution" case will be obvious when computing as we will see. Solve a triangle where = 95°, b = 4 and c = 5 We have SSA. We know an angle and a side opposite it so we'll use the Law of Sines. 4 a 95 5 We have the answer to sine and want to know the angle so we can use inverse sine. What happens when you put this in your calculator? Remember the domain of the inverse sine function is numbers from -1 to 1 since the sine values range from -1 to 1. What this means is there is no solution. (You can't build a triangle like this).

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**Solve a triangle where = 35°, b = 6 and c = 8**

The smallest angle should have the smallest side opposite it and the largest angle should have the largest side opposite it. We have SSA again so we know it could be the weird one of no solution, one solution or two solutions. 6 49.9 a 10.42 35 95.1 8 Knowing and , can you find ? Now how can you find a? Since this was an SSA triangle we need to check to see if there are two solutions. Remember your calculator only gives you one answer on the unit circle that has the sine value of You need to figure out where the other one is and see if you can make a triangle with it.

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**Looking at the same problem: Solve a triangle where = 35°, b = 6 and c = 8**

Let's check to see if there is another triangle possible. We got 49.9° from the calculator. Draw a picture and see if there is another angle whose sine is 6 130.1 a 2.69 = 130.1 35 14.9 49.9 So there IS another triangle. (remember our picture is not drawn to scale). 8 Knowing and , can you find ? Now how can you find a?

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**Solve a triangle where = 42°, b = 22 and c = 12**

22 21.4 a 29.40 42 116.6 Since this is SSA we need to check the other possible sine value for possibility of a second triangle solution. 12 Knowing and , can you find ?

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**Not possible to build another triangle with these stipulations.**

Solve a triangle where = 42°, b = 22 and c = 12 Not possible to build another triangle with these stipulations. 22 158.6 a 42 = 158.6 21.4 12 Knowing and , can you find ? This negative number tells us that there is no second triangle so this is the one triangle solution.

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Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar

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