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

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 78 , $\angle B= 18 ^{\circ}$ , and $\angle C= 66 ^{\circ}$ .

Possible Answers:

$\\BC = 169.82 \\AC = 30.38 \\\angle A = 96^\circ$

$\\BC = 254.73 \\AC = 36.38 \\\angle A = 96^\circ$

$\\BC = 80.91 \\AC = 21.104 \\\angle A = 96^\circ$

$\\BC = 87.91 \\AC = 7.914 \\\angle A = 96^\circ$

$\\BC = 84.91 \\AC = 26.38 \\\angle A = 96^\circ$

Correct answer:

$\\BC = 84.91 \\AC = 26.38 \\\angle A = 96^\circ$

Explanation:

The first step is to create a picture

Plot2

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 18 - 66 \\\angle A = 96^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 18 for B , 66 for A and 78 for a , and solve for the missing side.

$\\\left(\frac{\sin( 66 )}{ 78 }\right)= \left(\frac{\sin( 18 )}{ AC }\right) \\\\AC = \left(\frac{ 78 \sin( 18 )}{\sin( 66 )}\right) \\\\AC = 26.3843745919819$

Remember to round to the nearest hundredth.

AC = 26.38

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 695.9043999999999 + 6084 - 2 \cdot 26.38 \cdot 78 \cdot \cos( 96 ) \\BC^{2} = 7210.06829431611 \\BC = 84.9121210094066 \\BC = 84.91$

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Example Question #14 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 95 , $\angle B= 32 ^{\circ}$ , and $\angle C= 82 ^{\circ}$ .

Possible Answers:

$\\BC = 83.64 \\AC = 40.672 \\\angle A = 66^\circ$

$\\BC = 262.92 \\AC = 60.84 \\\angle A = 66^\circ$

$\\BC = 90.64 \\AC = 15.252 \\\angle A = 66^\circ$

$\\BC = 87.64 \\AC = 50.84 \\\angle A = 66^\circ$

$\\BC = 175.28 \\AC = 54.84 \\\angle A = 66^\circ$

Correct answer:

$\\BC = 87.64 \\AC = 50.84 \\\angle A = 66^\circ$

Explanation:

The first step is to create a picture

Plot3

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 32 - 82 \\\angle A = 66^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 32 for B , 82 for A and 95 for a , and solve for the missing side.

$\\\left(\frac{\sin( 82 )}{ 95 }\right)= \left(\frac{\sin( 32 )}{ AC }\right) \\\\AC = \left(\frac{ 95 \sin( 32 )}{\sin( 82 )}\right) \\\\AC = 50.8370730019896$

Remember to round to the nearest hundredth.

AC = 50.84

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 2584.7056000000002 + 9025 - 2 \cdot 50.84 \cdot 95 \cdot \cos( 66 ) \\BC^{2} = 7680.792322545 \\BC = 87.6401296356013 \\BC = 87.64$

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Example Question #14 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 182 , $\angle B= 3 ^{\circ}$ , and $\angle C= 81 ^{\circ}$ .

Possible Answers:

$\\BC = 366.52 \\AC = 13.64 \\\angle A = 96^\circ$

$\\BC = 183.26 \\AC = 9.64 \\\angle A = 96^\circ$

$\\BC = 549.78 \\AC = 19.64 \\\angle A = 96^\circ$

$\\BC = 186.26 \\AC = 2.892 \\\angle A = 96^\circ$

$\\BC = 179.26 \\AC = 7.712 \\\angle A = 96^\circ$

Correct answer:

$\\BC = 183.26 \\AC = 9.64 \\\angle A = 96^\circ$

Explanation:

The first step is to create a picture

Plot4

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 3 - 81 \\\angle A = 96^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 3 for B , 81 for A and 182 for a , and solve for the missing side.

$\\\left(\frac{\sin( 81 )}{ 182 }\right)= \left(\frac{\sin( 3 )}{ AC }\right) \\\\AC = \left(\frac{ 182 \sin( 3 )}{\sin( 81 )}\right) \\\\AC = 9.64387615477605$

Remember to round to the nearest hundredth.

AC = 9.64

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 92.92960000000001 + 33124 - 2 \cdot 9.64 \cdot 182 \cdot \cos( 96 ) \\BC^{2} = 33583.7157964677 \\BC = 183.25860360831 \\BC = 183.26$

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Example Question #81 : Similarity, Right Triangles, & Trigonometry

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 131 , $\angle B= 29 ^{\circ}$ , and $\angle C= 108 ^{\circ}$ .

Possible Answers:

$\\BC = 187.88 \\AC = 70.78 \\\angle A = 43^\circ$

$\\BC = 89.94 \\AC = 53.424 \\\angle A = 43^\circ$

$\\BC = 93.94 \\AC = 66.78 \\\angle A = 43^\circ$

$\\BC = 96.94 \\AC = 20.034 \\\angle A = 43^\circ$

$\\BC = 281.82 \\AC = 76.78 \\\angle A = 43^\circ$

Correct answer:

$\\BC = 93.94 \\AC = 66.78 \\\angle A = 43^\circ$

Explanation:

The first step is to create a picture

Plot5

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 29 - 108 \\\angle A = 43^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 29 for B , 108 for A and 131 for a , and solve for the missing side.

$\\\left(\frac{\sin( 108 )}{ 131 }\right)= \left(\frac{\sin( 29 )}{ AC }\right) \\\\AC = \left(\frac{ 131 \sin( 29 )}{\sin( 108 )}\right) \\\\AC = 66.7784292143583$

Remember to round to the nearest hundredth.

AC = 66.78

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 4459.5684 + 17161 - 2 \cdot 66.78 \cdot 131 \cdot \cos( 43 ) \\BC^{2} = 8824.54074913841 \\BC = 93.9390267627806 \\BC = 93.94$

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Example Question #82 : Similarity, Right Triangles, & Trigonometry

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 116 , $\angle B= 27 ^{\circ}$ , and $\angle C= 120 ^{\circ}$ .

Possible Answers:

$\\BC = 218.85 \\AC = 70.81 \\\angle A = 33^\circ$

$\\BC = 68.95 \\AC = 48.648 \\\angle A = 33^\circ$

$\\BC = 75.95 \\AC = 18.243 \\\angle A = 33^\circ$

$\\BC = 218.85 \\AC = 70.81 \\\angle A = 33^\circ$

$\\BC = 145.9 \\AC = 64.81 \\\angle A = 33^\circ$

$\\BC = 68.95 \\AC = 48.648 \\\angle A = 33^\circ$

$\\BC = 145.9 \\AC = 64.81 \\\angle A = 33^\circ$

$\\BC = 75.95 \\AC = 18.243 \\\angle A = 33^\circ$

$\\BC = 72.95 \\AC = 60.81 \\\angle A = 33^\circ$

Correct answer:

$\\BC = 72.95 \\AC = 60.81 \\\angle A = 33^\circ$

Explanation:

The first step is to create a picture

Plot6

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 27 - 120 \\\angle A = 33^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 27 for B , 120 for A and 116 for a , and solve for the missing side.

$\\\left(\frac{\sin( 120 )}{ 116 }\right)= \left(\frac{\sin( 27 )}{ AC }\right) \\\\AC = \left(\frac{ 116 \sin( 27 )}{\sin( 120 )}\right) \\\\AC = 60.8098766383251$

Remember to round to the nearest hundredth.

AC = 60.81

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 3697.8561000000004 + 13456 - 2 \cdot 60.81 \cdot 116 \cdot \cos( 33 ) \\BC^{2} = 5321.95882107139 \\BC = 72.9517568059289 \\BC = 72.95$

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Example Question #23 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 111 , $\angle B= 27 ^{\circ}$ , and $\angle C= 123 ^{\circ}$ .

Possible Answers:

$\\BC = 62.17 \\AC = 48.072 \\\angle A = 30^\circ$

$\\BC = 132.34 \\AC = 64.09 \\\angle A = 30^\circ$

$\\BC = 69.17 \\AC = 18.027 \\\angle A = 30^\circ$

$\\BC = 198.51 \\AC = 70.09 \\\angle A = 30^\circ$

$\\BC = 66.17 \\AC = 60.09 \\\angle A = 30^\circ$

Correct answer:

$\\BC = 66.17 \\AC = 60.09 \\\angle A = 30^\circ$

Explanation:

The first step is to create a picture

Plot7

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 27 - 123 \\\angle A = 30^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 27 for B , 123 for A and 111 for a , and solve for the missing side.

$\\\left(\frac{\sin( 123 )}{ 111 }\right)= \left(\frac{\sin( 27 )}{ AC }\right) \\\\AC = \left(\frac{ 111 \sin( 27 )}{\sin( 123 )}\right) \\\\AC = 60.0866983976108$

Remember to round to the nearest hundredth.

AC = 60.09

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 3610.8081 + 12321 - 2 \cdot 60.09 \cdot 111 \cdot \cos( 30 ) \\BC^{2} = 4379.04653402366 \\BC = 66.174364628787 \\BC = 66.17$

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Example Question #24 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 100 , $\angle B= 15 ^{\circ}$ , and $\angle C= 97 ^{\circ}$ .

Possible Answers:

$\\BC = 89.41 \\AC = 20.864 \\\angle A = 68^\circ$

$\\BC = 93.41 \\AC = 26.08 \\\angle A = 68^\circ$

$\\BC = 280.23 \\AC = 36.08 \\\angle A = 68^\circ$

$\\BC = 96.41 \\AC = 7.824 \\\angle A = 68^\circ$

$\\BC = 186.82 \\AC = 30.08 \\\angle A = 68^\circ$

Correct answer:

$\\BC = 93.41 \\AC = 26.08 \\\angle A = 68^\circ$

Explanation:

The first step is to create a picture

Plot8

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 15 - 97 \\\angle A = 68^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 15 for B , 97 for A and 100 for a , and solve for the missing side.

$\\\left(\frac{\sin( 97 )}{ 100 }\right)= \left(\frac{\sin( 15 )}{ AC }\right) \\\\AC = \left(\frac{ 100 \sin( 15 )}{\sin( 97 )}\right) \\\\AC = 26.0762730956666$

Remember to round to the nearest hundredth.

AC = 26.08

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 680.1664 + 10000 - 2 \cdot 26.08 \cdot 100 \cdot \cos( 68 ) \\BC^{2} = 8726.2184087426 \\BC = 93.4142302261417 \\BC = 93.41$

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Example Question #25 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 109 , $\angle B= 16 ^{\circ}$ , and $\angle C= 43 ^{\circ}$ .

Possible Answers:

$\\BC = 132.99 \\AC = 35.24 \\\angle A = 121^\circ$

$\\BC = 139.99 \\AC = 13.215 \\\angle A = 121^\circ$

$\\BC = 136.99 \\AC = 44.05 \\\angle A = 121^\circ$

$\\BC = 410.97 \\AC = 54.05 \\\angle A = 121^\circ$

$\\BC = 273.98 \\AC = 48.05 \\\angle A = 121^\circ$

Correct answer:

$\\BC = 136.99 \\AC = 44.05 \\\angle A = 121^\circ$

Explanation:

The first step is to create a picture

Plot9

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 16 - 43 \\\angle A = 121^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 16 for B , 43 for A and 109 for a , and solve for the missing side.

$\\\left(\frac{\sin( 43 )}{ 109 }\right)= \left(\frac{\sin( 16 )}{ AC }\right) \\\\AC = \left(\frac{ 109 \sin( 16 )}{\sin( 43 )}\right) \\\\AC = 44.0535836204938$

Remember to round to the nearest hundredth.

AC = 44.05

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 1940.4024999999997 + 11881 - 2 \cdot 44.05 \cdot 109 \cdot \cos( 121 ) \\BC^{2} = 18767.2616295538 \\BC = 136.993655435402 \\BC = 136.99$

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Example Question #26 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 99 , $\angle B= 2 ^{\circ}$ , and $\angle C= 79 ^{\circ}$ .

Possible Answers:

$\\BC = 99.61 \\AC = 3.52 \\\angle A = 99^\circ$

$\\BC = 298.83 \\AC = 13.52 \\\angle A = 99^\circ$

$\\BC = 102.61 \\AC = 1.056 \\\angle A = 99^\circ$

$\\BC = 95.61 \\AC = 2.816 \\\angle A = 99^\circ$

$\\BC = 199.22 \\AC = 7.52 \\\angle A = 99^\circ$

Correct answer:

$\\BC = 99.61 \\AC = 3.52 \\\angle A = 99^\circ$

Explanation:

The first step is to create a picture

Plot10

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 2 - 79 \\\angle A = 99^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 2 for B , 79 for A and 99 for a , and solve for the missing side.

$\\\left(\frac{\sin( 79 )}{ 99 }\right)= \left(\frac{\sin( 2 )}{ AC }\right) \\\\AC = \left(\frac{ 99 \sin( 2 )}{\sin( 79 )}\right) \\\\AC = 3.51971729370085$

Remember to round to the nearest hundredth.

AC = 3.52

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 12.3904 + 9801 - 2 \cdot 3.52 \cdot 99 \cdot \cos( 99 ) \\BC^{2} = 9922.41896475444 \\BC = 99.6113395390025 \\BC = 99.61$

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Example Question #191 : High School: Geometry

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 47 , $\angle B= 16 ^{\circ}$ , and $\angle C= 78 ^{\circ}$ .

Possible Answers:

$\\BC = 95.86 \\AC = 17.24 \\\angle A = 86^\circ$

$\\BC = 47.93 \\AC = 13.24 \\\angle A = 86^\circ$

$\\BC = 143.79 \\AC = 23.24 \\\angle A = 86^\circ$

$\\BC = 50.93 \\AC = 3.972 \\\angle A = 86^\circ$

$\\BC = 43.93 \\AC = 10.592 \\\angle A = 86^\circ$

Correct answer:

$\\BC = 47.93 \\AC = 13.24 \\\angle A = 86^\circ$

Explanation:

The first step is to create a picture

Plot11

Since we have 2 angles, we can solve for the 3rd angle easily.

$\\\angle A =180- 16 - 78 \\\angle A = 86^\circ$

Now we can use the Law of Sines to figure out one side

Recall the Law of Sines

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

Now plug in 16 for B , 78 for A and 47 for a , and solve for the missing side.

$\\\left(\frac{\sin( 78 )}{ 47 }\right)= \left(\frac{\sin( 16 )}{ AC }\right) \\\\AC = \left(\frac{ 47 \sin( 16 )}{\sin( 78 )}\right) \\\\AC = 13.2443771407098$

Remember to round to the nearest hundredth.

AC = 13.24

Now we need to solve for the last side of the triangle.

To solve for this, we can use the Law of Cosines.

Lets recall the Law of Cosines.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 175.29760000000002 + 2209 - 2 \cdot 13.24 \cdot 47 \cdot \cos( 86 ) \\BC^{2} = 2297.48148303701 \\BC = 47.9320506867483 \\BC = 47.93$

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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

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

Example Question #14 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #14 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #81 : Similarity, Right Triangles, & Trigonometry

Example Question #82 : Similarity, Right Triangles, & Trigonometry

Example Question #23 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #24 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #25 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #26 : Apply Laws Of Sines And Cosines: Ccss.Math.Content.Hsg Srt.D.11

Example Question #191 : High School: Geometry

All Common Core: High School - Geometry Resources

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