<< Chapter < Page Chapter >> Page >
In this section, you will:
  • Use the Law of Sines to solve oblique triangles.
  • Find the area of an oblique triangle using the sine function.
  • Solve applied problems using the Law of Sines.

Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation    measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in [link] that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles .

A diagram of a triangle where the vertices are the first ground station, the second ground station, and the airplane in the air between them. The angle between the first ground station and the plane is 15 degrees, and the angle between the second station and the airplane is 35 degrees. The side between the two stations is of length 20 miles. There is a dotted line perpendicular to the ground side connecting the airplane vertex with the ground - an altitude line.

Using the law of sines to solve oblique triangles

In any triangle, we can draw an altitude    , a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles.

Any triangle that is not a right triangle is an oblique triangle    . Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:

  1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See [link] .
    An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma's values are known, as is the side opposite beta, between alpha and gamma.
  2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See [link] .
    An oblique triangle consisting of angles alpha, beta, and gamma. Alpha and gamma are known, as is the side opposite alpha, between beta and gamma.
  3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See [link] .
    An oblique triangle consisting of angles alpha, beta, and gamma. Alpha is the only angle known. Two sides are known. The first is opposite alpha, between beta and gamma, and the second is opposite gamma, between alpha and beta.

Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in [link] .

An oblique triangle consisting of sides a, b, and c, and angles alpha, beta, and gamma. Side c is opposide angle gamma and is the horizontal base of the triangle. Side b is opposite angle beta, and side a is opposite angle alpha. There is a dotted perpendicular line - an altitude - from the gamma angle to the horizontal base c.

Using the right triangle relationships, we know that sin α = h b and sin β = h a . Solving both equations for h gives two different expressions for h .

h = b sin α  and  h = a sin β

We then set the expressions equal to each other.

            b sin α = a sin β   ( 1 a b ) ( b sin α ) = ( a sin β ) ( 1 a b ) Multiply both sides by 1 a b .                sin α a = sin β b

Similarly, we can compare the other ratios.

sin α a = sin γ c  and  sin β b = sin γ c

Collectively, these relationships are called the Law of Sines .

sin α a = sin β b = sin λ c

Note the standard way of labeling triangles: angle α (alpha) is opposite side a ; angle β (beta) is opposite side b ; and angle γ (gamma) is opposite side c . See [link] .

While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified.

A triangle with standard labels.

Law of sines

Given a triangle with angles and opposite sides labeled as in [link] , the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines    is based on proportions and is presented symbolically two ways.

sin α a = sin β b = sin γ c
a sin α = b sin β = c sin γ

To solve an oblique triangle, use any pair of applicable ratios.

Questions & Answers

difference between calculus and pre calculus?
Asma Reply
give me an example of a problem so that I can practice answering
Jenefa Reply
x³+y³+z³=42
Robert
dont forget the cube in each variable ;)
Robert
of she solves that, well ... then she has a lot of computational force under her command ....
Walter
what is a function?
CJ Reply
I want to learn about the law of exponent
Quera Reply
explain this
Hinderson Reply
what is functions?
Angel Reply
A mathematical relation such that every input has only one out.
Spiro
yes..it is a relationo of orders pairs of sets one or more input that leads to a exactly one output.
Mubita
Is a rule that assigns to each element X in a set A exactly one element, called F(x), in a set B.
RichieRich
If the plane intersects the cone (either above or below) horizontally, what figure will be created?
Feemark Reply
can you not take the square root of a negative number
Sharon Reply
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
Elaine Reply
can I get some pretty basic questions
Ama Reply
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
Amara Reply
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
Mars Reply
what is the domain of f(x)=x-4/x^2-2x-15 then
Conney Reply
x is different from -5&3
Seid
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
jeric Reply
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
jeric Reply
What are the question marks for?
Elliott
Practice Key Terms 4

Get the best Precalculus course in your pocket!





Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Precalculus' conversation and receive update notifications?

Ask