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Common Core: High School - Geometry : Similarity, Right Triangles, & Trigonometry

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 #71 : Similarity, Right Triangles, & Trigonometry

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:

5.71

0.53

0.63

1.58

1.90

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.

Report an Error

Example Question #71 : Similarity, Right Triangles, & Trigonometry

In a triangle where the side opposite a $45 ^{\circ}$ has length 13 find the side opposite a $65 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

0.06

0.81

1.23

15.99

10.57

Correct answer:

15.99

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 45 for , 13 for and 65 for .
Now our equation becomes

$\left(\frac{\sin( 45 )}{ 13 }\right)= \left(\frac{\sin( 65 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 13 \sin( 65 )}{\sin( 45 )}\right) \\\\b = -15.9863244465377$

Now we round our answer to the nearest tenth.

b = -15.99

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

b = 15.99

Report an Error

Example Question #72 : Similarity, Right Triangles, & Trigonometry

In a triangle where the side opposite a $26 ^{\circ}$ has length 12 find the side opposite a $41 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

15.58

1.30

9.24

0.77

0.06

Correct answer:

15.58

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 26 for , 12 for and 41 for .
Now our equation becomes

$\left(\frac{\sin( 26 )}{ 12 }\right)= \left(\frac{\sin( 41 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 12 \sin( 41 )}{\sin( 26 )}\right) \\\\b = -15.5778376306642$

Now we round our answer to the nearest tenth.

b = -15.58

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

b = 15.58

Report an Error

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

In a triangle where the side opposite a $36 ^{\circ}$ has length 6 find the side opposite a $58 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

0.62

3.7

9.73

0.27

1.62

Correct answer:

3.7

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 36 for , 6 for and 58 for .
Now our equation becomes

$\left(\frac{\sin( 36 )}{ 6 }\right)= \left(\frac{\sin( 58 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 6 \sin( 58 )}{\sin( 36 )}\right) \\\\b = -3.69854363983686$

Now we round our answer to the nearest tenth.

b = -3.7

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

b = 3.7

Report an Error

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

In a triangle where the side opposite a $26 ^{\circ}$ has length 11 find the side opposite a $17 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

0.25

4.06

0.37

44.63

2.71

Correct answer:

2.71

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 26 for , 11 for and 17 for .
Now our equation becomes

$\left(\frac{\sin( 26 )}{ 11 }\right)= \left(\frac{\sin( 17 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 11 \sin( 17 )}{\sin( 26 )}\right) \\\\b = 2.71142149312156$

Now we round our answer to the nearest tenth.

b = 2.71

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

Report an Error

Example Question #12 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

In a triangle where the side opposite a $81 ^{\circ}$ has length 11 find the side opposite a $66 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

12.23

1.11

0.90

9.90

0.08

Correct answer:

12.23

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 81 for , 11 for and 66 for .
Now our equation becomes

$\left(\frac{\sin( 81 )}{ 11 }\right)= \left(\frac{\sin( 66 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 11 \sin( 66 )}{\sin( 81 )}\right) \\\\b = 12.2250984628794$

Now we round our answer to the nearest tenth.

b = 12.23

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

Report an Error

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

In a triangle where the side opposite a $70 ^{\circ}$ has length 5 find the side opposite a $50 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

13.29

0.38

1.88

0.53

2.66

Correct answer:

1.88

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 70 for , 5 for and 50 for .
Now our equation becomes

$\left(\frac{\sin( 70 )}{ 5 }\right)= \left(\frac{\sin( 50 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 5 \sin( 50 )}{\sin( 70 )}\right) \\\\b = -1.88083767684263$

Now we round our answer to the nearest tenth.

b = -1.88

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

b = 1.88

Report an Error

Example Question #12 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

In a triangle where the side opposite a $11 ^{\circ}$ has length 5 find the side opposite a $64 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

7.33

0.68

3.41

1.47

0.29

Correct answer:

3.41

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 11 for , 5 for and 64 for .
Now our equation becomes

$\left(\frac{\sin( 11 )}{ 5 }\right)= \left(\frac{\sin( 64 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 5 \sin( 64 )}{\sin( 11 )}\right) \\\\b = -3.41136278039312$

Now we round our answer to the nearest tenth.

b = -3.41

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

b = 3.41

Report an Error

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

In a triangle where the side opposite a $78 ^{\circ}$ has length 6 find the side opposite a $52 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

1.08

6.49

5.54

0.18

0.92

Correct answer:

5.54

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 78 for , 6 for and 52 for .
Now our equation becomes

$\left(\frac{\sin( 78 )}{ 6 }\right)= \left(\frac{\sin( 52 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 6 \sin( 52 )}{\sin( 78 )}\right) \\\\b = 5.54498774628774$

Now we round our answer to the nearest tenth.

b = 5.54

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

Report an Error

Example Question #72 : Similarity, Right Triangles, & Trigonometry

In a triangle where the side opposite a $44 ^{\circ}$ has length 7 find the side opposite a $53 ^{\circ}$ angle. Round you answer to the nearest hundredth.

Possible Answers:

1.05

6.67

7.34

0.95

0.15

Correct answer:

6.67

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 44 for , 7 for and 53 for .
Now our equation becomes

$\left(\frac{\sin( 44 )}{ 7 }\right)= \left(\frac{\sin( 53 )}{ b }\right)$

Now we rearrange the equation to solve for b

$\\b = \left(\frac{ 7 \sin( 53 )}{\sin( 44 )}\right) \\\\b = 6.67220075275393$

Now we round our answer to the nearest tenth.

b = 6.67

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

Report an Error

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All Common Core: High School - Geometry Resources

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Common Core: High School - Geometry : Similarity, Right Triangles, & Trigonometry

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

All Common Core: High School - Geometry Resources

Example Questions

Example Question #71 : Similarity, Right Triangles, & Trigonometry

Example Question #71 : Similarity, Right Triangles, & Trigonometry

Example Question #72 : Similarity, Right Triangles, & Trigonometry

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

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

Example Question #12 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

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

Example Question #12 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

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

Example Question #72 : Similarity, Right Triangles, & Trigonometry

All Common Core: High School - Geometry Resources

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