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Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.

  1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.
  2. Apply the Law of Cosines to find the length of the unknown side or angle.
  3. Apply the Law of Sines    or Cosines to find the measure of a second angle.
  4. Compute the measure of the remaining angle.

Finding the unknown side and angles of a sas triangle

Find the unknown side and angles of the triangle in [link] .

A triangle with standard labels. Side a = 10, side c = 12, and angle beta = 30 degrees.

First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines.

Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side b , as we know the measurement of the opposite angle β .

b 2 = a 2 + c 2 2 a c cos β b 2 = 10 2 + 12 2 2 ( 10 ) ( 12 ) cos ( 30 ) Substitute the measurements for the known quantities . b 2 = 100 + 144 240 ( 3 2 ) Evaluate the cosine and begin to simplify . b 2 = 244 120 3 b = 244 120 3 Use the square root property . b 6.013

Because we are solving for a length, we use only the positive square root. Now that we know the length b , we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle α , we have

sin α a = sin β b sin α 10 = sin ( 30° ) 6.013 sin α = 10 sin ( 30° ) 6.013 Multiply both sides of the equation by 10 . α = sin 1 ( 10 sin ( 30° ) 6.013 ) Find the inverse sine of  10 sin ( 30° ) 6.013 . α 56.3°

The other possibility for α would be α = 180° 56.3° 123.7°. In the original diagram, α is adjacent to the longest side, so α is an acute angle and, therefore, 123.7° does not make sense. Notice that if we choose to apply the Law of Cosines    , we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between and 180°. Proceeding with α 56.3° , we can then find the third angle of the triangle.

γ = 180° 30° 56.3° 93.7°

The complete set of angles and sides is

α 56.3° a = 10 β = 30° b 6.013 γ 93.7° c = 12
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Find the missing side and angles of the given triangle: α = 30° , b = 12 , c = 24.

a 14.9 , β 23.8° , γ 126.2° .

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Solving for an angle of a sss triangle

Find the angle α for the given triangle if side a = 20 , side b = 25 , and side c = 18.

For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle α , we have

               a 2 = b 2 + c 2 −2 b c cos α               20 2 = 25 2 + 18 2 −2 ( 25 ) ( 18 ) cos α Substitute the appropriate measurements .               400 = 625 + 324 900 cos α Simplify in each step .               400 = 949 900 cos α            −549 = −900 cos α Isolate cos  α .            −549 −900 = cos α             0.61 cos α cos −1 ( 0.61 ) α Find the inverse cosine .                   α 52.4°

See [link] .

A triangle with standard labels. Side b =25, side a = 20, side c = 18, and angle alpha = 52.4 degrees.
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Given a = 5 , b = 7 , and c = 10 , find the missing angles.

α 27.7° , β 40.5° , γ 111.8°

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Solving applied problems using the law of cosines

Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.

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
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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
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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 2

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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