Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Law of Cosines and Sines

Study concepts, example questions & explanations for Precalculus

Create An Account

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 Next →

Example Question #1 : Use The Laws Of Cosines And Sines

Which is NOT an angle of the following triangle?

(Not drawn to scale.)

Possible Answers:

$48.3^{\circ}$

$91.2^{\circ}$

$33.9^{\circ}$

$54.9^{\circ}$

Correct answer:

$48.3^{\circ}$

Explanation:

In order to solve this problem, we need to find the angles of the triangle. Only then will we be able to find which answer choice is NOT an angle. Using the Law of Cosines we are able to find each angle.

$\cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{9^{2}+11^{2}-6^{2}}{2\times 9\times 11}$

cos(A)=0.83

$A=\cos^{-1}(0.83)=33.9^{\circ}$

$\cos(B)=\frac{c^{2}+a^{2}-b^{2}}{2ac}=\frac{11^{2}+6^{2}-9^{2}}{2\times 6\times 11}$

cos(B)=0.575

$B=\cos^{-1}(0.575)=54.9^{\circ}$

To find angle we use the formula again or we can remember that the angles in a triangle add up to $180^{\circ}$ .

$C^{\circ}=180^{\circ}-33.9^{\circ}-54.9^{\circ}=91.2^{\circ}$

The answer choice that isn't an actual angle of the triangle is $48.3^{\circ}$ .

Report an Error

Example Question #41 : Trigonometric Applications

Solve the triangle

Law_of_cosines__sss_

Possible Answers:

$A = 99^{\circ}, B = 21^{\circ}, C=54^{\circ}$

None of the other answers

$A = 95^{\circ}, B = 23^{\circ}, C=62^{\circ}$

$A = 115^{\circ}, B = 22^{\circ}, C=43^{\circ}$

$A = 100^{\circ}, B = 31^{\circ}, C=49^{\circ}$

Correct answer:

$A = 115^{\circ}, B = 22^{\circ}, C=43^{\circ}$

Explanation:

Since we are given all 3 sides, we can use the Law of Cosines in the angle form:

$cos(A) = \frac{-a^2+b^2+c^2}{2bc}$

$cos(B) = \frac{a^2-b^2+c^2}{2ac}$

$cos(C) = \frac{a^2+b^2-c^2}{2ab}$

Let's start by finding angle A:

$cos(A) = \frac{-(12)^2+5^2+9^2}{2\cdot5\cdot9} = \frac{-38}{90} = -0.4\overline{22}$

$A = cos^{-1}(-0.4\overline{22}) = 115^{\circ}$

Now let's solve for B:

$cos(B) = \frac{12^2-(5)^2+9^2}{2\cdot12\cdot9} = \frac{200}{216} = 0.\overline{925}$

$B = cos^{-1}(0.\overline{925}) = 22^{\circ}$

We can solve for C the same way, but since we now have A and B, we can use our knowledge that all interior angles of a triangle must add up to 180 to find C.

$C = 180^{\circ} - 115^{\circ} - 22^{\circ} = 43^{\circ}$

Report an Error

Example Question #3 : Use The Laws Of Cosines And Sines

Given three sides of the triangle below, determine the angles $\theta$ , $\phi$ , and $\beta$ in degrees.

Sss

Possible Answers:

$\theta \approx 47.0^\circ , \phi \approx 74.9^\circ , \beta \approx 58.1^\circ$

$\theta \approx 75.5^\circ , \phi \approx 57.9^\circ , \beta \approx 46.6^\circ$

$\theta \approx 58.1^\circ , \phi \approx 47.0^\circ , \beta \approx 74.9^\circ$

$\theta \approx 45.0^\circ , \phi \approx 65.0^\circ , \beta \approx 70.0^\circ$

Correct answer:

$\theta \approx 75.5^\circ , \phi \approx 57.9^\circ , \beta \approx 46.6^\circ$

Explanation:

We are only given sides, so we must use the Law of Cosines. The equation for the Law of Cosines is

$a^2=b^2+c^2-2bc\cos{A}$ ,

where , and are the sides of a triangle and the angle is opposite the side .

We have three known sides and three unknown angles, so we must write the Law three times, where each equation lets us solve for a different angle.

To solve for angle $\theta$ , we write

$16^2=12^2+14^2-2(12)(14) \cos{\theta}$ and solve for $\theta$ using the inverse cosine function $\arccos$ on a calculator to get

$\theta \approx 75.5$ .

Similarly, for angle $\phi$ ,

$14^2=12^2+16^2-2(12)(16) \cos{\phi}$

$\phi \approx 57.9$

and for $\beta$ ,

$12^2=14^2+16^2-2(14)(16) \cos{\beta}$

and $\beta \approx 46.6$

Report an Error

Example Question #4 : Use The Laws Of Cosines And Sines

Lc2b

What is the measurement of $\angle C$ ? Round to the nearest tenth, if needed.

Possible Answers:

$84^\circ$

$83^\circ$

$83.8^\circ$

$83.7^\circ$

Correct answer:

$83.8^\circ$

Explanation:

We need to use the Law of Cosines for side then solve for cosC .

Therefore,

$\frac{c^2-a^2-b^2}{-2ab}=cosC$ .

Plugging in the information provided, we have:

$\frac{50^2-44^2-29^2}{-2(44)(29)}=cosC$ .

Then simplify, 0.1085423197=cosC .

To solve for , use $cos^{-1}(.1085423197)=C$ .

Solve and then round to the appropriate units: 83.76870649=83.8 . Therefore, $C=83.8^\circ$ .

Report an Error

Example Question #5 : Use The Laws Of Cosines And Sines

Find the measure, in degrees, of the largest angle in a triangle whose sides measure , , and .

Possible Answers:

$102.8^\circ$

$93.8^\circ$

$119.8^\circ$

$87.8^\circ$

$111.8^\circ$

Correct answer:

$111.8^\circ$

Explanation:

When all three sides are given, Law of Cosines is appropriate.

Since 10 is the largest side length, it is opposite the largest angle and thus should be the c-value in the equation below.

$\\* c^2 = a^2 + b^2 - 2ab \cos (C) \\ \\* 10^2 = 5^2 + 7^2 - 2 \cdot 5 \cdot 7 \cos \(C) \\\\* C = \arccos \left(\frac {10^2-5^2-7^2}{2 \cdot 5 \cdot 7}\right) \approx 111.8^\circ$

Report an Error

Example Question #6 : Use The Laws Of Cosines And Sines

Use the Law of Cosines to find .

(Triangle not drawn to scale.)

Possible Answers:

b=15.2

b=6.6

b=135

b=8

b=12

Correct answer:

b=6.6

Explanation:

We need to use the Law of Cosines in order to solve this problem

$b^{2}=a^{2}+c^{2}-2ac\cos(B)$

in this case, $a=6, c=10, B=40^{\circ}$

In order to arrive at our answer, we plug the numbers into our formula:

$b^{2}=6^{2}+10^{2}-2(6)(10)\cos (40^{\circ})$

$b^{2}=36+100-91.9$

$b\approx6.6$

Note: we use the "approximately" to indicate the answer is around 6.6. It will vary depending on your rounding.

Report an Error

Example Question #7 : Use The Laws Of Cosines And Sines

Use the Law of Cosines to find .

(Triangle not drawn to scale.)

Possible Answers:

a=74

a=131.5

a=93

a=17

a=9.6

Correct answer:

a=9.6

Explanation:

In order to solve this problem, we need to use the following formula

$a^{2}=b^{2}+c^{2}-2bc\cos(A)$

in this case, $b=11, c=7, A=60^{\circ}$

We plug our numbers into our formula and get our answer:

$a^{2}=11^{2}+7^{2}-2(11)(7)\cos (60^{\circ})$

$a^{2}=121+49-77$

$a^{2}=93$

$a\approx 9.6$

Note: we use the "approximately" to indicate that the answer is around 9.6. It will vary depending on your rounding.

Report an Error

Example Question #1 : Use The Laws Of Cosines And Sines

The 2 sides of a triangle have lengths of 10 and 20. The included angle is 25 degrees. What is the length of the third side to the nearest integer?

Possible Answers:

Correct answer:

Explanation:

Write the formula for the Law of Cosines.

c^2=a^2+b^2-2abcos(C)

Substitute the side lengths of the triangle and the included angle to find the third length.

c^2=10^2+20^2-2(10)(20)cos(25)

c^2=100+400-400cos(25)

c^2=500-400(.9063)

c^2=137.48

c=11.725

Round this to the nearest integer.

c=12

Report an Error

Example Question #9 : Use The Laws Of Cosines And Sines

What is the approximate length of the unknown side of the triangle if two sides of the triangle are and , with an included angle of $30^\circ$ ?

Possible Answers:

12.9

12.4

13.4

20.5

Correct answer:

12.4

Explanation:

Write the formula for the Law of Cosines.

c^2=a^2+b^2-2abcos(C)

Substitute the known values and solve for .

c^2=10^2+20^2-2(10)(20)cos(30)

$c^2=100+400-400(\frac{\sqrt3}{2})$

$c^2=500-200\sqrt3$

$c=\sqrt{500-200\sqrt3} =12.3931=12.4$

Report an Error

Example Question #10 : Use The Laws Of Cosines And Sines

Lc1c

What is the measurement of side using the Law of Cosines? Round to the nearest tenth.

Possible Answers:

235m

234.3m

234.2m

234m

Correct answer:

234.3m

Explanation:

The Law of Cosines for side is,

$a^2=b^2+c^2-2bc\cdot cosA$ .

Plugging in the information we know, the formula is,

a^2=400^2+375^2-2(400)(375)cos(35)=54879.38671 .

Then take the square of both sides: a=234.2634985 .

Finally, round to the appropriate units: a=234.3m .

Report an Error

← Previous 1 2 3 4 Next →

View Pre-Calculus Tutors

Michael
Certified Tutor

University of Ghana Legon, Bachelor in Arts, Statistics. University of Louisiana-Monroe, Non Degree Bachelors, Education.

View Pre-Calculus Tutors

Andrew
Certified Tutor

University of Wisconsin-Stevens Point, Bachelor of Science, Mathematics. University of Wisconsin-Oshkosh, Master of Science, ...

View Pre-Calculus Tutors

Susana I
Certified Tutor

California State Polytechnic University-Pomona, Bachelor of Science, Geology. The University of Texas at El Paso, Master of S...

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GRE Tutors in Houston, LSAT Tutors in Chicago, LSAT Tutors in Phoenix, SSAT Tutors in Los Angeles, MCAT Tutors in Miami, Algebra Tutors in New York City, Calculus Tutors in Chicago, ACT Tutors in Houston, MCAT Tutors in Los Angeles, English Tutors in Phoenix

Popular Courses & Classes

LSAT Courses & Classes in Miami, MCAT Courses & Classes in Chicago, SAT Courses & Classes in New York City, SSAT Courses & Classes in Los Angeles, GMAT Courses & Classes in Phoenix, Spanish Courses & Classes in Atlanta, SSAT Courses & Classes in Boston, GMAT Courses & Classes in Atlanta, MCAT Courses & Classes in San Diego, ACT Courses & Classes in Atlanta

Popular Test Prep

SSAT Test Prep in Phoenix, GMAT Test Prep in Phoenix, ISEE Test Prep in Washington DC, SSAT Test Prep in Seattle, SAT Test Prep in New York City, GRE Test Prep in Atlanta, SSAT Test Prep in Los Angeles, SAT Test Prep in Phoenix, ACT Test Prep in San Diego, ISEE Test Prep in San Francisco-Bay Area

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

Precalculus : Law of Cosines and Sines

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Use The Laws Of Cosines And Sines

Example Question #41 : Trigonometric Applications

Example Question #3 : Use The Laws Of Cosines And Sines

Example Question #4 : Use The Laws Of Cosines And Sines

Example Question #5 : Use The Laws Of Cosines And Sines

Example Question #6 : Use The Laws Of Cosines And Sines

Example Question #7 : Use The Laws Of Cosines And Sines

Example Question #1 : Use The Laws Of Cosines And Sines

Example Question #9 : Use The Laws Of Cosines And Sines

Example Question #10 : Use The Laws Of Cosines And Sines

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: