Call Now to Set Up Tutoring:

(888) 888-0446

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

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

Create An Account

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 #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.

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

Report an Error

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.

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

Report an Error

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 .

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

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

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

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

$\implies cos(x) = 0.875$

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

$\implies x = 28.96$

Report an Error

Example Question #91 : Similarity, Right Triangles, & Trigonometry

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:

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

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

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

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

Correct answer:

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

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.

Report an Error

Example Question #92 : Similarity, Right Triangles, & Trigonometry

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:

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

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

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

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

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.

Report an Error

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

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

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

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

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

$\implies 29.1676 = C^2$

$\implies 5.40 = C$

Report an Error

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

Report an Error

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

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

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

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

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

$\implies 18.7939 = B(0.9848)$

$\implies 19.08 = B$

Report an Error

Example Question #1 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles

Using the sides of a right triangle, what is the definition of $tan(\theta)$ ?

Possible Answers:

$tan(\theta)=\frac{adjacent}{hypotenuse}$

$tan(\theta)=\frac{opposite}{adjacent}$

$tan(\theta)=\frac{opposite}{hypotenuse}$

Correct answer:

$tan(\theta)=\frac{opposite}{adjacent}$

Explanation:

We can see this to be true by looking at the following triangle.

We are considering $\angle C$ . The opposite side to $\angle C$ is $\bar{AB}$ . $\angle C$ has two adjacent sides, but since $\bar{AC}$ is opposite to the right angle, this is the hypotenuse of the triangle. So the adjacent side to $\angle C$ must be $\bar{BC}$ . We can set up the following equation to check and make sure that $tan(\theta)=\frac{opposite}{adjacent}$ .

$tan(\theta)=\frac{opposite}{adjacent}$

$tan(36.87) = \frac{3}{4}$

0.75 = 0.75

This shows that $tan(\theta)=\frac{opposite}{adjacent}$ .

Report an Error

Example Question #1 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles

True or False: By similarity, side ratios in right triangles are properties of the angles in the triangle.

Possible Answers:

False

True

Correct answer:

True

Explanation:

Similar right triangles are two right triangles that differ in side lengths but have congruent corresponding angles. This means that if you have an angle, $\theta_1$ , in the first triangle and an angle, $\theta_2$ , in the second triangle. So $\theta_1 = \theta_2$ .If we are considering the cosine of these two angles.

$cos(\theta_1) = \cos(\theta_2)$

$\frac{adjacent_1}{hypotenuse_1} = \frac{adjacent_2}{hypotenuse_2}$

Side ratios would also follow from computing the sine and tangent of the angles using their sides as well. This shows that the side ratios are properties of the angles in the triangles.

Report an Error

View Tutors

Caroline
Certified Tutor

University of Florida, Master's/Graduate, Religion.

View Tutors

Stephanie
Certified Tutor

The University of Texas at San Antonio, Bachelor in Arts, Multi-/Interdisciplinary Studies, General.

View Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

All Common Core: High School - Geometry Resources

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

Popular Subjects

SSAT Tutors in Denver, French Tutors in Dallas Fort Worth, GRE Tutors in Seattle, ACT Tutors in Los Angeles, ISEE Tutors in New York City, English Tutors in Philadelphia, GRE Tutors in San Francisco-Bay Area, Spanish Tutors in San Diego, Statistics Tutors in San Diego, Chemistry Tutors in Miami

Popular Courses & Classes

GMAT Courses & Classes in Los Angeles, GRE Courses & Classes in San Diego, SAT Courses & Classes in Houston, ACT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Denver, GMAT Courses & Classes in Denver, SAT Courses & Classes in Denver, SSAT Courses & Classes in New York City, ISEE Courses & Classes in Boston, Spanish Courses & Classes in Dallas Fort Worth

Popular Test Prep

LSAT Test Prep in Philadelphia, SAT Test Prep in Houston, ACT Test Prep in Dallas Fort Worth, ISEE Test Prep in Philadelphia, MCAT Test Prep in Miami, ACT Test Prep in San Francisco-Bay Area, GMAT Test Prep in New York City, GMAT Test Prep in Phoenix, ISEE Test Prep in Seattle, MCAT Test Prep in Houston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

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 #191 : High School: Geometry

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 #91 : Similarity, Right Triangles, & Trigonometry

Example Question #92 : Similarity, Right Triangles, & Trigonometry

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

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

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

Example Question #1 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles

Example Question #1 : Understand That By Similarity, Side Ratios In Right Triangles Are Properties Of The Angles In The Triangle, Leading To Definitions Of Trigonometric Ratios For Acute Angles

All Common Core: High School - Geometry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: