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

To the nearest hundredth, solve $\bigtriangleup ABC$ where, AB = 187 , $\angle B= 29 ^{\circ}$ , and $\angle C= 55 ^{\circ}$ .

Possible Answers:

$\\BC = 230.03 \\AC = 33.201 \\\angle A = 96^\circ$

$\\BC = 681.09 \\AC = 120.67 \\\angle A = 96^\circ$

$\\BC = 454.06 \\AC = 114.67 \\\angle A = 96^\circ$

$\\BC = 223.03 \\AC = 88.536 \\\angle A = 96^\circ$

$\\BC = 227.03 \\AC = 110.67 \\\angle A = 96^\circ$

Correct answer:

$\\BC = 227.03 \\AC = 110.67 \\\angle A = 96^\circ$

Explanation:

The first step is to create a picture

Plot12

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

$\\\angle A =180- 29 - 55 \\\angle A = 96^\circ$

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 29 for B , 55 for A and 187 for a , and solve for the missing side.

$\\\left(\frac{\sin( 55 )}{ 187 }\right)= \left(\frac{\sin( 29 )}{ AC }\right) \\\\AC = \left(\frac{ 187 \sin( 29 )}{\sin( 55 )}\right) \\\\AC = 110.674690514574$

Remember to round to the nearest hundredth.

AC = 110.67

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.

$\\BC^{2} = a^{2} - 2 a b \cos{\left (BC \right )} + b^{2} \\BC^2 = 12247.8489 + 34969 - 2 \cdot 110.67 \cdot 187 \cdot \cos( 96 ) \\BC^{2} = 51543.3426211569 \\BC = 227.03158947855 \\BC = 227.03$

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Example Question #93 : Similarity, Right Triangles, & Trigonometry

Using the Law of Cosines, solve for . Round to the second decimal place.

Screen shot 2020 08 07 at 2.49.50 pm

Possible Answers:

x = 45

x = 15.13

x = 28.96

x = 67.89

Correct answer:

x = 28.96

Explanation:

We will begin by defining the variables in our equation $A^2 + B^2 -2ABcos(\gamma) = C^2$ .

We will let

A = 16

B = 12

C = 8

$\gamma = x$

Next we plug our values into our formulas and solve for .

$A^2 + B^2 -2ABcos(\gamma) = C^2$

$\implies 16^2 + 12^2 - 2(16)(12)cos(x) = 8^2$

$\implies 256 + 144 - 384cos(x) = 64$

$\implies -384cos(x) = -336$

$\implies cos(x) = 0.875$

$\implies x = cos^{-1}(0.875)$

$\implies x = 28.96$

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Example Question #201 : High School: Geometry

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

Screen shot 2020 08 13 at 8.30.42 am

Possible Answers:

$A^2 + B^2 -2ABcos(\gamma) = C^2$

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

$A^2 + B^2 -2ABsin(\gamma) = C^2$

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

Correct answer:

$A^2 + B^2 -2ABsin(\gamma) = C^2$

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.

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Example Question #202 : High School: Geometry

Which of the following expressions properly states the Law of Sines? Use the following triangle for reference.

Screen shot 2020 08 13 at 8.30.42 am

Possible Answers:

$A^2 + B^2 - 2ABsin(\gamma) = C^2$

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

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

$A^2 + B^2 -2ABcos(\gamma) = C^2$

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.

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Example Question #96 : Similarity, Right Triangles, & Trigonometry

Use the Law of Cosine to find the missing side. Round to the second decimal place.

Screen shot 2020 08 20 at 8.44.42 am

Possible Answers:

C = 3.57

C = 2.53

8.79

C = 5.40

Correct answer:

C = 5.40

Explanation:

We are able to use our equation from the Law of Cosine A^2 + B^2 - 2ABcos(c) = C^2 . We will start by assigning the values in our figure to the variables in the formula.

We will let:

A = 3

B = 4

c = 100

Now we can plug these values into our formula.

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

$\implies 3^2 + 4^2 - (2)(3)(4)cos(100) = C^2$

$\implies 9 + 16 - 24cos(100) = C^2$

$\implies 25 - (24)(-0.1736) = C^2$ $\implies 25 - (24)(-0.1736) = C^2$

$\implies 25 - (-4.1676) = C^2$

$\implies 29.1676 = C^2$

$\implies 5.40 = C$

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Example Question #97 : Similarity, Right Triangles, & Trigonometry

Using the Law of Sines, solve for and . Round to the second decimal place.

Screen shot 2020 08 20 at 10.01.08 am

Possible Answers:

x = 6, y = 6

x = 9.02, y =8.46

x = 6.72, y = 7.83

x = 10.29, y = 5.32

Correct answer:

x = 9.02, y =8.46

Explanation:

We begin by labeling our sides and angles according the variables in the equation for the Law of Sines: $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$

We will let

A = 6

B = x B = x

C = y

$\alpha = 75$

$\beta = 65$

$\gamma = 40$

We will begin by solving for . From our equation $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ , we will leave out the term to solve for since it is not needed at this time.

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

$\implies \frac{6}{sin(40)} = \frac{x}{sin(75)}$

$\implies (6)sin(75) = (x)sin(40)$ (by cross-multiplication)

$\implies (6)(0.9659) = (x)(0.6428)$

$\implies 5.7954 = (x)(0.6428)$

$\implies 9.02 = x$

Now we will solve for . We could use $\frac{B}{sin(\alpha)} = \frac{C}{sin(\beta)}$ for our relationship, but we can also use $\frac{A}{sin(\gamma)} = \frac{C}{sin(\beta)}$ .

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

$\implies \frac{6}{sin(40)} = \frac{y}{sin(65)}$

$\implies (6)sin(65) = (y)sin(40)$ (by cross-multiplication)

$\implies 5.4378 = (y)(0.6428)$

$\implies 8.46 = y$

So x = 9.02, y =8.46

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Example Question #93 : Similarity, Right Triangles, & Trigonometry

Use the Law of Sines to find the missing side lengths and angles. Round to the second decimal place.

Screen shot 2020 08 20 at 11.17.26 am

Possible Answers:

$C = 19.08, \alpha = 90, B = 10.15$

$C = 9.17, \alpha = 70, B = 15.63$

$C = 10.15, \alpha = 70, B = 19.08$

$C = 12.13, \alpha = 45, B = 23.45$

Correct answer:

$C = 10.15, \alpha = 70, B = 19.08$

Explanation:

We know for the Law of Sines our equation is $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)} = \frac {C}{sin(\beta)}$ . We begin by assigning our values on the figure to variables in the equation.

We will let:

A = 20

$\gamma = 80$

$\beta = 30$

We will begin by solving for side .

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

$\implies \frac{20}{sin(80)} = \frac{C}{sin(30)}$

$\implies (20)sin(30) = Csin(80)$

$\implies (20)(0.5) = C(0.9848)$

$\implies 10.15 = C$

Using the Law of Sines we need to solve for $\alpha$ and using the relation $\frac{A}{sin(\gamma)} = \frac{B}{sin(\alpha)}$ . We do not know either or $\alpha$ . We are able to solve for $\alpha$ because we know that the angles of a triangle must add up to 180.

$30 + 80 + \alpha = 180$

$30 + 80 + \alpha = 180$ 70 = 180

Now we have enough information to solve for using the Law of Sines

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

$\implies \frac{20}{sin(80)} = \frac{B}{sin(70)}$

$\implies 20sin(70) = Bsin(80)$

$\implies (20)(0.9397) = B(0.9848)$

$\implies 18.7939 = B(0.9848)$

$\implies 19.08 = B$

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Example Question #1 : Circles

Given $\frac{C}{C'}=25$ , and R=4 , what is R'?

Screen shot 2020 07 01 at 7.06.26 pm

Possible Answers:

$R'=\frac{4}{25}$

R'=25

R'=4

$R'=\frac{25}{4}$

R'=100

Correct answer:

$R'=\frac{4}{25}$

Explanation:

In order to do this problem, we need to remember the relationship between Circumference's, and Radii. $\frac{C}{C'}=\frac{R}{R'}$ . With this relationship, we can determine what is.

$\frac{C}{C'}=\frac{R}{R'}$

Substitute for the variables given in the question.

$25=\frac{4}{R'}$

Multiply by on each side.

25R'=4

Divide by to get .

$R'=\frac{4}{25}$

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Example Question #1 : Circles

Given $\frac{C}{C'}=118$ , and R=12 , what is R'?

Screen shot 2020 07 01 at 7.06.26 pm

Possible Answers:

R'=118

$R'=\frac{6}{59}$

$R'=\frac{1}{118}$

R'=6

R'=59

Correct answer:

$R'=\frac{6}{59}$

Explanation:

In order to do this problem, we need to remember the relationship between Circumference's, and Radii. $\frac{C}{C'}=\frac{R}{R'}$ . With this relationship, we can determine what is.

$\frac{C}{C'}=\frac{R}{R'}$

Substitute for the variables given in the question.

$118=\frac{12}{R'}$

Multiply by on each side.

118R'=12

Divide by 118 to get .

$R'=\frac{12}{118}=\frac{6}{59}$

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Example Question #2 : Circles

Given $\frac{C}{C'}=115$ , and R=9 , what is R'?

Screen shot 2020 07 01 at 7.06.26 pm

Possible Answers:

$R'=\frac{9}{115}$

R'=106

$R'=\frac{115}{9}$

R'=115

R'=9

Correct answer:

$R'=\frac{9}{115}$

Explanation:

In order to do this problem, we need to remember the relationship between Circumference's, and Radii. $\frac{C}{C'}=\frac{R}{R'}$ . With this relationship, we can determine what is.

$\frac{C}{C'}=\frac{R}{R'}$

Substitute for the variables given in the question.

$115=\frac{9}{R'}$

Multiply by on each side.

115R'=9

Divide by 115 to get .

$R'=\frac{9}{115}$

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

Example Question #93 : Similarity, Right Triangles, & Trigonometry

Example Question #201 : High School: Geometry

Example Question #202 : High School: Geometry

Example Question #96 : Similarity, Right Triangles, & Trigonometry

Example Question #97 : Similarity, Right Triangles, & Trigonometry

Example Question #93 : Similarity, Right Triangles, & Trigonometry

Example Question #1 : Circles

Example Question #1 : Circles

Example Question #2 : Circles

All Common Core: High School - Geometry Resources

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