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Common Core: High School - Geometry : High School: Geometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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

Choose the answer choice that correctly states the Law of Sines, given the triangle below.

Screen shot 2020 08 19 at 4.54.09 pm

Possible Answers:

$\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$

$A^{2}+B^{2}-2AB\sin (\gamma )=C^{2}$

$A^{2}+B^{2}-2AB\cos (\gamma )=C^{2}$

$\frac{A}{\cos (\gamma )}=\frac{B}{\cos (\alpha )}=\frac{C}{\cos (\beta )}$

Correct answer:

$\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$

Explanation:

The Law of Sines uses ratios of a triangle’s sides and angles to allow us to be able to solve for unknown sides and/or angles. This relationship is true for any triangle.

Screen shot 2020 08 19 at 4.54.09 pm

The Law of Sines is $\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$ .

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

Which of the following expressions correctly states the Law of Cosines? Use the following triangle for reference.

Screen shot 2020 08 20 at 10.56.17 am

Possible Answers:

$\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$

$A^{2}+B^{2}-2AB\sin (\gamma )=C^{2}$

$A^{2}+B^{2}-2AB\cos (\gamma )=C^{2}$

$\frac{A}{\cos (\gamma )}=\frac{B}{\cos (\alpha )}=\frac{C}{\cos (\beta )}$

Correct answer:

$A^{2}+B^{2}-2AB\cos (\gamma )=C^{2}$

Explanation:

The Law of Cosines should look similar to a theorem we use quite commonly for right triangles. The Law of Cosines is basically the Pythagorean Theorem, but it has an added argument so that it is true for any and all triangles.

Screen shot 2020 08 20 at 10.56.17 am

The Law of Cosines is $A^{2}+B^{2}-2AB\cos (\gamma )=C^{2}$

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

Choose the answer choice that correctly states the Law of Sines, given the triangle below.

Screen shot 2020 08 20 at 11.04.58 am

Possible Answers:

$A^{2}+B^{2}-2AB\cos (\gamma )=C^{2}$

$A^{2}+B^{2}-2AB\sin (\gamma )=C^{2}$

$\frac{A}{\cos (\gamma )}=\frac{B}{\cos (\alpha )}=\frac{C}{\cos (\beta )}$

$\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$

Correct answer:

$\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$

Explanation:

The Law of Sines uses ratios of a triangle’s sides and angles to allow us to be able to solve for unknown sides and/or angles. This relationship is true for any triangle.

Screen shot 2020 08 20 at 11.04.58 am

The Law of Sines is $\frac{A}{\sin (\gamma )}=\frac{B}{\sin (\alpha )}=\frac{C}{\sin (\beta )}$ .

Report an Error

Example Question #1 : 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.

Possible Answers:

1.32

7.95

4.59

3.78

Correct answer:

4.59

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 A^2 + B^2 - 2ABcos(c) = C^2 . We will begin by defining our variables according to the problem.

We will let

A = 6

B = 6

c = 45

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.

A^2 + B^2 - 2ABcos(c) = C^2

$\implies 6^2 + 6^2 - (2)(6)(6)cos(45) = C^2$

$\implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$

$\implies 72 - 36\sqrt{2} = C^2$

$\implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$ $\implies 36 + 36 - 72(\frac{1}{\sqrt{2}}) = C^2$ $\implies 21.0883 = C^2$

$\implies 4.59 = C$

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Example Question #2 : 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.

Possible Answers:

Just the Law of Sines

Just the Law of Cosines

Neither the Law of Cosines and Law of Sines

Both the Law of Cosines and Law of Sines

Correct answer:

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

$60 + 60 + \gamma = 180$

$\implies \gamma = 60$

Now we have enough information to use either law. We will begin by using the law of cosine A^2 + B^2 - 2ABcos(c) = C^2 .

We will let

A=7

B=7

C=60

Now we will plug these values into our formula

A^2 + B^2 - 2ABcos(c) = C^2

$\implies 7^2 + 7^2 - (2)(7)(7)cos(60) = C^2$

$\implies 49 + 49 - 98(\frac{1}{2})= C^2$

$\implies 98 - 49 = C^2$

$\implies 49 = C^2$

$\implies 7 = C^2$

We can also prove that using the Law of Sines will give us the same answer. Our formula for the Law of Sines is $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ .

We will let:

A=7

C=7

$\alpha = 60$

$\beta = 60$

$\gamma = 60$

We can set up with the relation $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

$\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$

$\implies \frac{7}{sin(60)} = \frac{B}{sin(60)}$

$\implies 7sin(60) = Bsin(60)$ (by cross-multiplication)

$\implies \frac{7sin(60)}{sin(60)} = B$

$\implies 7 = B$ ( sin(60) cancels)

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Example Question #61 : Similarity, Right Triangles, & Trigonometry

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.

Possible Answers:

False

True

Correct answer:

True

Explanation:

Just looking at the formula A^2 + B^2 - 2ABcos(c) = C^2 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.

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Example Question #62 : Similarity, Right Triangles, & Trigonometry

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.

Possible Answers:

True

False

Correct answer:

False

Explanation:

Consider the following relationship: $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$ . Let’s say we know side A = 9 and $\alpha = 60$ . Plugging this into our relationship we have $\frac{9}{sin(\gamma)} = \frac{B}{sin(60)}$ . 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.

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

In a triangle where the side opposite a $62 ^{\circ}$ has length 10 find the side opposite a $66 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

11.36

0.88

0.09

1.14

8.81

Correct answer:

11.36

Explanation:

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

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

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

$\left(\frac{\sin( 62 )}{ 10 }\right)= \left(\frac{\sin( 66 )}{ b }\right)$

Now we rearrange the equation to solve for

$\\b = \left(\frac{ 10 \sin( 66 )}{\sin( 62 )}\right) \\\\b = -11.3557732263804$

Now we round our answer to the nearest tenth.

b = -11.36

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

b = 11.36

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Example Question #2 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

In a triangle where the side opposite a $41 ^{\circ}$ has length 6 find the side opposite a $39 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

5.58

1.08

6.45

0.93

0.15

Correct answer:

6.45

Explanation:

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

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

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

$\left(\frac{\sin( 41 )}{ 6 }\right)= \left(\frac{\sin( 39 )}{ b }\right)$

Now we rearrange the equation to solve for

$\\b = \left(\frac{ 6 \sin( 39 )}{\sin( 41 )}\right) \\\\b = 6.45402256283484$

Now we round our answer to the nearest tenth.

b = 6.45

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

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Example Question #3 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

In a triangle where the side opposite a $60 ^{\circ}$ has length 3 find the side opposite a $51 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

1.90

0.53

5.71

0.63

1.58

Correct answer:

1.58

Explanation:

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

$\frac{1}{A} \sin{\left (a \right )} = \frac{1}{B} \sin{\left (b \right )}$

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

$\left(\frac{\sin( 60 )}{ 3 }\right)= \left(\frac{\sin( 51 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 3 \sin( 51 )}{\sin( 60 )}\right) \\\\b = 1.57596947141406$

Now we round our answer to the nearest tenth.

b = 1.58

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

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All Common Core: High School - Geometry Resources

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Common Core: High School - Geometry : High School: Geometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

All Common Core: High School - Geometry Resources

Example Questions

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

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

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

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

Example Question #2 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #61 : Similarity, Right Triangles, & Trigonometry

Example Question #62 : Similarity, Right Triangles, & Trigonometry

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

Example Question #2 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #3 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

All Common Core: High School - Geometry Resources

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