Common Core: High School - Geometry : Apply Laws of Sines and Cosines: CCSS.Math.Content.HSG-SRT.D.11

Example Questions

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Example Question #5 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Your friend wants to build a triangular mirror with a wooden frame for her living room.  She already has two cut wooden pieces that are 6 inches in length.  She has put these sides together at an angle of 45 degrees.  What length should she cut the third side to be?  Round to the second decimal place.

Explanation:

Since we know two sides and their adjoining angle, we are able to use the Law of Cosine to solve for the third side.  Recall that the formula for the Law of Cosine is .  We will begin by defining our variables according to the problem.

We will let

How do we know that  is the angle we should use?   is the angle that would be directly opposite of side  (our unknown side) so this is our desired angle to use in the formula.  Now we will plug these values into our formula.

Example Question #6 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

You are given a triangle with 2 known sides and 2 known angles.  Which method could you use to solve for the unknown side and angle.

Neither the Law of Cosines and Law of Sines

Just the Law of Cosines

Both the Law of Cosines and Law of Sines

Just the Law of Sines

Both the Law of Cosines and Law of Sines

Explanation:

Take the triangle below for example:

First, we can figure out the third angle simply because we know all three angles of a triangle must add up to be 180.  So

Now we have enough information to use either law.  We will begin by using the law of cosine .

We will let

Now we will plug these values into our formula

We can also prove that using the Law of Sines will give us the same answer.  Our formula for the Law of Sines is .

We will let:

We can set up with the relation

(by cross-multiplication)

( cancels)

Example Question #9 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

True or False: You are looking at an obtuse triangle.  You are given two side lengths and their adjoining angle.  You are able to figure out the third side length using the Law of Cosines.

True

False

True

Explanation:

Just looking at the formula   we can see that this is similar to the Pythagorean Theorem.  We commonly use the Pythagorean Theorem to solve for side , which is what this problem is asking us to do.  Consider the picture below.  We are able to solve for the third side using the information from the picture.  This is the same information given in the question.

Example Question #10 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

True or False: We are given the length of one side and one angle of a triangle. This is enough information to use the Law of Sines to solve for the other sides and angles.

True

False

False

Explanation:

Consider the following relationship: .  Let’s say we know side and .  Plugging this into our relationship we have .  We still have two unknowns and no other information to solve for these.  This is not enough information to use the Law of Sines to solve for the unknown sides and angles.

Example Question #1 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 10 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

In order to solve this, we need to recall the law of sines.

Where , and  are angles, and , and , are opposite side lengths.

Now let's plug in 62 for , 10 for  and 66 for .
Now our equation becomes

Now we rearrange the equation to solve for

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Example Question #1 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

In a triangle where the side opposite a  has length 6 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

In order to solve this, we need to recall the law of sines.

Where , and  are angles, and , and , are opposite side lengths.

Now let's plug in 41 for , 6 for  and 39 for .
Now our equation becomes

Now we rearrange the equation to solve for

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Example Question #3 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 3 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

In order to solve this, we need to recall the law of sines.

Where , and  are angles, and , and , are opposite side lengths.

Now let's plug in 60 for , 3 for  and 51 for .
Now our equation becomes

Now we rearrange the equation to solve for b

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Example Question #4 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 13 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

In order to solve this, we need to recall the law of sines.

Where , and  are angles, and , and , are opposite side lengths.

Now let's plug in 45 for , 13 for  and 65 for .
Now our equation becomes

Now we rearrange the equation to solve for b

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Example Question #5 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 12 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

In order to solve this, we need to recall the law of sines.

Where , and  are angles, and , and , are opposite side lengths.

Now let's plug in 26 for , 12 for  and 41 for .
Now our equation becomes

Now we rearrange the equation to solve for b

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

Example Question #6 : Prove Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.10

In a triangle where the side opposite a  has length 6 find the side opposite a  angle. Round you answer to the nearest hundredth.

Explanation:

In order to solve this, we need to recall the law of sines.

Where , and  are angles, and , and , are opposite side lengths.

Now let's plug in 36 for , 6 for  and 58 for .
Now our equation becomes

Now we rearrange the equation to solve for b

Now we round our answer to the nearest tenth.

Remember if your answer is negative, multiply it by -1, because side lengths can't be negative.

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