Law of Sines and Cosines - SSA Case

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En provenance du cours de University of California, Irvine

Pre-Calculus: Trigonometry

251 notes

University of California, Irvine

Pre-Calculus: Trigonometry

251 notes

This course covers mathematical topics in trigonometry. Trigonometry is the study of triangle angles and lengths, but trigonometric functions have far reaching applications beyond simple studies of triangles. This course is designed to help prepare students to enroll for a first semester course in single variable calculus. Upon completing this course, you will be able to: 1. Evaluate trigonometric functions using the unit circle and right triangle approaches 2. Solve trigonometric equations 3. Verify trigonometric identities 4. Prove and use basic trigonometric identities. 5. Manipulate trigonometric expressions using standard identities 6. Solve right triangles 7. Apply the Law of Sines and the Law of Cosines

À partir de la leçon

Law of Sines and Law of Cosines

The Law of Sines and the Law of Cosines give useful properties of the trigonometry functions that can help us solve for unknown angles and sides in oblique (non-right angle) triangles. We will focus on utilizing those laws in solving triangles, including those which arise in word problems.

Law of Sines and Cosines - ASA Case 6:04

Law of Sines and Cosines - SAS Case 5:32

Law of Sines and Cosines - SSA Case 10:04

Law of Sines and Cosines - SSS Case 10:29

Law of Sines Word Problem 3:20

Law of Cosines Word Problem 4:09

Rencontrer les enseignants

Dr. Sarah Eichhorn
Associate Dean, Distance Learning & Tenured Lecturer
Dr. Rachel Cohen Lehman

Let's work with the law of Sines and Cosines.

For example, the triangle ABC be an oblique triangle with little b equal to 37,

little a equal to 54 and capital B equal to 30 degrees.

Let's solve the triangle,

that is we're going to find all the missing sides and angles.

So, let's mark on our figure here the given information.

We have that b is equal to 37,

a is equal to 54 and capital B is 30 degrees.

Now, this is what we call the SSA case,

stands for Side-Side-Angle,

because we're given two sides and an angle that's not included between them.

Now, since we're given a matching angle and side pair,

namely capital B and little b,

we can apply the Law of Sines.

And since we're given little a,

we can use the Law of Sines to find capital A,

namely sine of A divided by a is

equal to sine of B divided by little b,

or sine of A is equal to a times sign of B divided by

b which is equal to 54 times the sine of 30 degrees divided by 37.

And the sine of 30 degrees is 1/2,

so this is 54 times 1/2 divided by 37 which is

equal to 27 divided by 37 which is approximately 0.7297.

Now, since A is an angle in a triangle,

its degree measure must be between zero and 180 degrees.

And therefore there's two possibilities for A.

Let's call them A1 and A2.

A1 is inverse sine of this 27 divided by 37

which is approximately 46.8612 degrees and there's

also A2 which is a 180 degrees minus A1 which

is approximately a 133.1388 degrees.

Now the question is, do they both work?

Well, they'll both work as long as when we add B

to these measures we're still under 180 degrees.

So, let's compute.

A1 plus B which is approximately this

46.8612 plus 30 degrees which is 76.8612,

which is less than 180 degrees.

So, yes. A1 will work. What about A2?

A2 plus B is approximately is 133.1388

degrees plus 30 degrees which is

equal to 163.1388 which is also less than 180 degrees,

which means A2 will also work,

which means there's two solutions.

We can form two triangles with this given information.

Now, how this turned out to be greater than 180 degrees?

Then, there would only be one solution and also sometimes in these cases,

we get no solution.

Coming back over here on the left.

If this turned would've turned out to be greater than one for example,

then there would be no solution because there is

no angle A whose sine is larger than one.

But in this case, we will have two solutions.

The first solution is when is A is equal to

A1 or this 46.8612 degrees.

So, if we know A and we know B then we can find C

because the angle measures in a triangle have to add up to 180 degrees,

that is C is equal to 180 minus A plus B,

which is approximately equal to 180 degrees minus 46.8612

degrees plus 30 degrees, which is

equal to 103.1388 degrees.

And now that we know capital C,

we can find little c by using a Law of Sines again,

that is C divided by sine of capital C is equal to B divided by sine of capital B

or C is equal to B times sine of C divided by sine of B

which is approximately 37 times sine of 103.1388 degrees

divided by the sine of

30 degrees, which using our calculator we get that this is approximately 72.0629.

Alright, and what about our second solution?

When A is equal

to A2 or this 133.1388 degrees.

Again, we can find C by using the fact that

the angles in the triangle add up to 180 degrees

that is C is equal to 180 degrees minus A plus B,

which is approximately 180 degrees minus 133.1388

degrees plus 30 degrees which

is equal to 16.8612 degrees.

And again, we can use the Law of Sines to find little c,

that is C is equal to B times sine of capital C divided by sine of capital B,

which is approximately 37 times sine of 16.8612

degrees divided by sine of

30 degrees, which is approximately 21.4641.

So, let's round our answers to the nearest tenth and write them up here.

So, there's two solutions.

The first solution, in addition to this information here

given in the problem set capital

A is equal to 46.9 degrees,

capital C is equal to 103.1

degrees and little c is equal to 72.1.

And the second solution is that A is

equal to 133.1 degrees,

capital C is equal to 16.9 degrees and little c is equal

to 21.5 and this

is an example of how we can solve triangles in the SSA Case.

Thank you and we'll see you next time.

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