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Law of Sines

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Law of Sines

When given a right triangle, you can use basic trigonometry to solve for missing information. When given SSS or SAS, you can use the Law of Cosines to solve for the missing information. But what happens when you are given two sides of a triangle and an angle that is not included? There are many ways to show two triangles are congruent, but SSA is not one of them. Why not? 

Watch This

http://www.youtube.com/watch?v=APNkWrD-U1k Khan Academy: Proof: Law of Sines

http://www.youtube.com/watch?v=dxYVBbSXYkA James Sousa: The Law of Sines: The Basics

Guidance

When given two sides and an angle that is not included between the two sides, you can use the Law of Sines. The Law of Sines states that in every triangle the ratio of each side to the sine of its corresponding angle is always the same. Essentially, it clarifies the general concept that opposite the largest angle is always the longest side. 

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Here is a proof of the Law of Sines:

Looking at the right triangle formed on the left:

\sin A & = \frac{h}{b}\\h & = b \sin A

Looking at the right triangle formed on the right:

\sin B & = \frac{h}{a}\\h & = a \sin B

Equating the heights which must be identical:

a \sin B & = b \sin A\\\frac{a}{\sin A} & = \frac{b}{\sin B}

The best way to use the Law of Sines is to draw an extremely consistent picture each and every time even if that means redrawing and relabeling a picture. The reason why the consistency is important is because sometimes given SSA information defines zero, one or even two possible triangles. 

Always draw the given angle in the bottom left with the two given sides above. 

In this image side  a is deliberately too short, but in most problems you will not know this. You will need to compare a to the height. 

\sin A & = \frac{h}{c}\\h & = c \ \sin A

Case 1: a < h

Simply put, side a  is not long enough to reach the opposite side and the triangle is impossible. 

Case 2: a = h

Side a  just barely reaches the opposite side forming a 90^\circ  angle. 

Case 3: h < a < c

In this case side  a can swing toward the interior of the triangle or the exterior of the triangle- there are two possible triangles. This is called the ambiguous case because the given information does not uniquely identify one triangle. To solve for both triangles, use the Law of Sines to solve for angle C_1  first and then use the supplement to determine C_2 .

Case 4:  c \le a

In this case, side a  can only swing towards the exterior of the triangle, only producing C_1 .

Example A

\angle A = 40^\circ, c = 13 , and a = 2 . If possible, find \angle C .

Solution:

\sin 40^\circ & = \frac{h}{13}\\h & = 13 \sin 40^\circ \approx 8.356

Because a < h \ (2 < 8.356) , this information does not form a proper triangle.

Example B

\angle A = 17^\circ, c = 14 , and a = 4.0932 \ldots   If possible, find \angle C .

Solution:

  \sin 17^\circ & = \frac{h}{14}\\h & = 14 \ \sin 17^\circ \approx 4.0932 \ldots

Since a = h , this information forms exactly one triangle and angle  C must be 90^\circ .

Example C

\angle A = 22^\circ, c = 11  and a = 9 . If possible, find \angle C .

Solution:

\sin 22^\circ & = \frac{h}{11}\\h & = 11 \ \sin 22^\circ \approx 4.12 \ldots

Since h < a < c , there must be two possible angles for angle C

Apply the Law of Sines:

\frac{9}{\sin 22^\circ} & = \frac{11}{\sin C_1}\\9 \sin C_1 & = 11 \sin 22^\circ\\\sin C_1 & = \frac{11 \sin 22^\circ}{9}\\C_1 & = \sin^{-1} \left( \frac{11 \sin 22^\circ}{9} \right) \approx 27.24 \ldots ^\circ\\C_2 & = 180 - C_1 = 152.75 \ldots^\circ

Concept Problem Revisited

SSA is not a method from Geometry that shows two triangles are congruent because it does not always define a unique triangle. 

Vocabulary

Ambiguous means that the given information may not uniquely identify one triangle. 

Guided Practice

1. Given  \Delta ABC where A = 10^\circ, b = 10, a = 11 , find \angle B

2. Given  \Delta ABC where A = 12^\circ, B = 50^\circ, a = 14  find b .

3. Given  \Delta ABC where A = 70^\circ, b = 8, a = 3 , find  \angle B if possible. 

Answers:

1.  \frac{10}{\sin B} = \frac{11}{\sin 10^\circ}

B = \sin^{-1} \left( \frac{10 \sin 10^\circ}{11} \right) \approx 9.08 \ldots ^\circ

2.  \frac{14}{\sin 12^\circ} = \frac{b}{\sin 50^\circ}

b = \frac{14 \sin 50^\circ}{\sin 12^\circ} \approx 51.58 \ldots

3. \sin 70^\circ = \frac{h}{8}

h = 8 \sin 70^\circ \approx 7.51 \ldots

Because a < h , this triangle is impossible. 

Practice

For 1-3, draw a picture of the triangle and state how many triangles could be formed with the given values.

1.  A = 30^\circ, a = 13, b = 15

2.  A = 22^\circ, a = 21, b = 12

3.  A = 42^\circ, a = 36, b = 37

For 4-7, find all possible measures of \angle B  (if any exist) for each of the following triangle values.

4. A = 86^\circ, a =15, b = 11

5.  A = 30^\circ, a = 24, b = 43

6.  A = 48^\circ, a = 34, b = 39

7.  A = 80^\circ, a = 22, b = 20

For 8-12, find the length of  b for each of the following triangle values.

8. A = 94^\circ, a = 31, B = 34^\circ

9.  A = 112^\circ, a = 12, B = 15^\circ

10.  A = 78^\circ, a = 20, B = 16^\circ

11. A = 54^\circ, a = 15, B = 112^\circ

12. A = 39^\circ, a = 9, B = 98^\circ

13. In  \Delta ABC, b=10 and \angle A = 39^\circ . What's a possible value for a  that would produce two triangles?

14. In  \Delta ABC, b=10 and \angle A = 39^\circ . What's a possible value for a  that would produce no triangles?

15. In  \Delta ABC, b=10 and \angle A = 39^\circ . What's a possible value for a  that would produce one triangle?

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