Download Module 6 Lesson 1 Law of Sines Part 1 Notes (1)

Module 6 Lesson 1 Solving Triangles using the Law of Sines – Part 1 We spent time in module 4 solving triangles, which were all right triangles. Now we learn what to do if the triangles we’re considering are NOT right triangles. Example 1: Solve this triangle: a = 15 cm, A = 25o, C = 100o In the last module, you solved many triangles but they were all right triangles. We don’t have a right triangle in this example. Let’s sketch our picture: The above picture is not drawn to scale. The important thing is that side “a” is opposite Angle A, side “b” is opposite Angle B, and side “c” is opposite Angle C. Finding Angle B should be fairly easy. Once again, all three angles of the triangle need to add up to 180° . Simply subtract: 180 − 100 − 25 = 55 B = 55o Now, we address the more difficult problem of finding the missing sides. The three trig functions (sine, cosine, tangent) and methods we learned previously only apply to right triangles. We need something new to help us with this problem: 1
Law of Sines That new helpful tool is called the Law of Sines. The Law of Sines applies to all triangles (including right triangles). Here it is: a
b
c
Law of Sines: =
=
sin A sin B sin C
The law states that the ratio between each side length and the sine of its opposite angle is the same for each side/angle combination in the triangle. Note that there are three sections to the Law of Sines. We use only two of those sections at one time. Let’s look back at our example. We have Angle A, Angle C and side “a” given to us. We can therefore use the Law of Sines like this: a
c
=
sin A sin C
Put in your values but be warned! The place where most students miss these problems is that they forget to put sin on the bottom of the fraction along with the angle. They simply put the angle measure. Don’t make this mistake: 15
c
15
c
RIGHT: WRONG: =
=
o
o
sin 25 sin100
25 100
15
c
=

We have a proportion, so we cross-­‐multiply to solve: sin 25 sin100


15 ⋅ sin100 = c ⋅sin 25
15 ⋅ sin100 c ⋅ sin 25
=
sin 25
sin 25
o
Divide each side by sin 25 : 15 ⋅ sin100
=c
sin 25
We’ve now solved for “c ”. Use your calculator and close your parenthesis around the numerator!!! This is second most popular place that students make mistakes on this problem. They don’t close their parentheses. You should get c ≈ 35.0 cm 2
Is your answer reasonable? 100 degrees is bigger than 25 degrees, so our answer should be longer than 15 cm. b
c
a
b
=
Now let’s find “b”. Should we use: or =
sin B sinC
sin A sin B
Does it matter? Either ratio will lead you to find “b” however, we found “c ” and angle “B” through the our calculations, make sure you didn’t make any mistakes in the calculations before you use either “c” or “B.” Example 1 continued a
b
b
c
=
=
sin A sin B
sin B sinC
15
b
b
35
=
=



sin 25 sin55
sin55 sin100
15 ⋅ sin55 = b ⋅ sin 25
35 ⋅ sin55 = b ⋅ sin100
o
r 15 ⋅ sin55 b ⋅ sin 25
35 ⋅ sin55 b ⋅ sin100
=
=
sin 25
sin 25
sin100
sin100
15 ⋅ sin55
35 ⋅ sin55
=
b
=b
sin 25
sin100
29.1 cm = b
29.1 cm = b
Our solution is: b ≈ 29.1 cm, c ≈ 35 cm, B = 55 ****Please note that we could have also solved this triangle by first calculating the 15
b
measurement of side “b” by using the ratio and then use the value of =
sin 25° sin 55°
c
29.1
c
15
“b” to find the measurement of “c ” using =
or
=
sin100° sin 55°
sin100° sin 25°
3
Example 2: Solve this triangle: b = 22 ft, A = 40o, B = 80o The first step you need to do is complete a sketch of the triangle and label what you are given. 180° − 80° − 40° = 60° therefore ∠ C = 60°
First I am going to solve for the missing angle
Next I am going to solve for "a" using the Law of Sines
Choose 2 ratios that you can use to solve for a missing piece
Substitute what you know in the ratio
b
a
=
sin B sin A
22
a
=
sin 80° sin 40°
Solve for the unknown piece my cross multiplying
22 ⋅sin 40° = asin 80°
Divide each side by sin 80°
22sin 40° asin 80°
=
sin 80°
sin 80°
a ≈ 14.4 ft
4
Next we are going to solve for the final side "c" by setting up another ratio
Choose 2 ratios that you can use to solve for a missing piece
Substitute what you know in the ratio
b
c
=
sin B sinC
22
c
=
sin 80° sin 60°
Solve for the unknown piece my cross multiplying
22 ⋅sin 60° = csin 80°
Divide each side by sin 80°
22sin 60° csin 80°
=
sin 80°
sin 80°
c ≈ 19.3 ft
Our solution is: C = 60°, a ≈ 14.4 ft, and c ≈ 19.3 ft Do you have to solve this triangle with these EXACT steps? Could you do the process differently or use different ratios? Different Situations with Triangles There’s more to the Law of Sines that we’ll get to in the next set of notes. I’d like to talk about something else first: What information do you have when given a triangle to solve? We know that you need to have three pieces of information (of a possible six) from a triangle in order to be able to fill in the missing parts. This is true. However, there’s a little more to it than that. The specific information you have will help indicate which method you should use to solve the triangle. 5
Here are possible combinations of three pieces of information you could be given: What you are given Diagram of Triangle Two angles and the side NOT in between them (Angle Angle Side) (AAS) Two angles and the side in between them (Angle Side Angle) (ASA) How do you solve for the missing pieces 1. Use the sum of the angles in a triangle add to be 180° to solve for the remaining parts. 2. Use the Law of Sines (You can complete these steps in reverse order if you choose) 1. Use the sum of the angles in a triangle add to be 180° to solve for the remaining parts. 2. Use the Law of Sines Two sides and an angle NOT in between them (Side Side Angle) (SSA) Two sides and the angle in between them (Side Angle Side) (SAS) Three sides (Side Side Side) (SSS) Three angles (Angle Angle Angle (AAA) 1. Check to see how many solutions the triangle has (this is only done on the SSA case) 2. Use the Law of Sines 3. Use the Law of Sines and the sum of the angles in a triangle add to be 180° to solve for the remaining parts. 1. Use Law of Cosines 2. Use the Law of Cosines finding the largest angle first. Then use the sum of the angles in a triangle add to be 180° to solve for the remaining parts 1. Use the Law of Cosines. 2. Use the Law of Cosines finding the largest angle first. Then use the sum of the angles in a triangle add to be 180° to solve for the remaining parts This triangle cannot be solved. http://www.mathsisfun.com/algebra/trig-­‐
solving-­‐triangles.html 6
In the 2 examples above we were given two angles and the side NOT in between them (AAS). As we move forward to look at the other combinations that we can be given, you need to start recognizing which of these situations you have and which method you’ll use to solve. Sometimes, you’ll use Law of Sines. Other times, you’ll start with the Law of Cosines (Lesson 2). You’ll have to learn the different situations. 7