Call Now to Set Up Tutoring:

(888) 888-0446

HiSET: Math : Problems involving right triangle trigonometry

Study concepts, example questions & explanations for HiSET: Math

Create An Account

All HiSET: Math Resources

125 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Sine

$\tan N = \frac{\sqrt{y}}{5}$

Evaluate $\sin N$ in terms of .

Possible Answers:

$\sqrt{ \frac{y}{y+25}}$

$\frac{5} {\sqrt{y}}$

$\frac{5}{ \sqrt{y+25}}$

$\sqrt{ \frac{y+25}{y}}$

$\frac{ \sqrt{y+25}}{5}$

Correct answer:

$\sqrt{ \frac{y}{y+25}}$

Explanation:

Suppose we allow O, A, H be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring .

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, so

$\tan N = \frac{O}{A} =\frac{\sqrt{y}}{5}$

We can set the lengths of the opposite and adjacent legs to $\sqrt{y}$ and 5, respectively. The length of the hypotenuse can be determined using the Pythagorean Theorem:

$H = \sqrt{O^{2}+A^{2}} = \sqrt{( \sqrt{y})^{2}+5^{2}} = \sqrt{y+25}$

The sine of the angle is equal to the ratio of the length opposite leg to that of the hypotenuse, so

$\sin N = \frac{O}{H} = \frac{\sqrt{y }}{ \sqrt{y+25}} =\sqrt{ \frac{y}{y+25}}$ .

Report an Error

Example Question #1 : Sine

$\cos N = \frac{x}{3}$

Evaluate $\sin N$ in terms of .

Possible Answers:

$\frac{3}{x}$

$\frac{x}{\sqrt{x^{2}-9}}$

$\frac{\sqrt{x^{2}-9}}{3}$

$\frac{3}{\sqrt{x^{2}-9}}$

$\frac{\sqrt{x^{2}-9}}{x}$

Correct answer:

$\frac{\sqrt{x^{2}-9}}{3}$

Explanation:

Suppose we allow O, A, H be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring . The cosine is defined to be the ratio of the length of the adjacent side to that of the hypotenuse, so

$\cos N = \frac{A}{H} = \frac{x}{3}$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = \sqrt{H^{2}- A^{2}} = \sqrt{x^{2}-3^{2}} = \sqrt{x^{2}-9 }$

The sine of the angle is equal to the ratio of the length of the opposite leg to that of the hypotenuse, so

$\sin N = \frac{O}{H} = \frac{\sqrt{x^{2}-9 }}{3}$ .

Report an Error

Example Question #1 : Problems Involving Right Triangle Trigonometry

$\tan N = \frac{\sqrt{y}}{5}$

Evaluate $\cos N$ in terms of .

Possible Answers:

$\sqrt{ \frac{y+25}{y}}$

$\frac{ \sqrt{y+25}}{5}$

$\frac{5}{ \sqrt{y+25}}$

$\frac{5} {\sqrt{y}}$

$\sqrt{ \frac{y}{y+25}}$

Correct answer:

$\frac{5}{ \sqrt{y+25}}$

Explanation:

Suppose we allow O, A, H be the lengths of the opposite leg and adjacent leg and the hypotenuse, respectively, of the right triangle with an acute angle measuring .

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, so

$\tan N = \frac{O}{A} =\frac{\sqrt{y}}{5}$

We can set the lengths of the opposite and adjacent legs to $\sqrt{y}$ and 5, respectively. The length of the hypotenuse can be determined using the Pythagorean Theorem:

$H = \sqrt{O^{2}+A^{2}} = \sqrt{( \sqrt{y})^{2}+5^{2}} = \sqrt{y+25}$

The cosine of the angle is equal to the ratio of the length of the adjacent leg to that of the hypotenuse, so

$\cos N = \frac{A}{H} = \frac{5}{ \sqrt{y+25}}$ .

Report an Error

Example Question #1 : Problems Involving Right Triangle Trigonometry

$\sin a^{\circ } = \frac{2}{5}$

$\cos (90-a)^{\circ } = ?$

Possible Answers:

$\frac{2}{\sqrt{29}}$

$\frac{5}{\sqrt{29}}$

$\frac{2}{\sqrt{21}}$

$\frac{5}{\sqrt{21}}$

None of the other choices gives the correct response.

Correct answer:

None of the other choices gives the correct response.

Explanation:

An identity of trigonometry is

$\sin a^{\circ} = \cos (90-a)^{\circ}$

for any value of .

Since $\sin a^{\circ } = \frac{2}{5}$ , it immediately follows that $\cos (90-a)^{\circ} =\frac{2}{5}$ .

This response is not among the given choices.

Report an Error

Example Question #1 : Problems Involving Right Triangle Trigonometry

Right

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

Possible Answers:

None of the other choices gives the correct response.

$\frac{144}{ \tan 35^{\circ }}$

$\frac{72}{ \tan 35^{\circ }}$

$72 \tan 35^{\circ }$

$144 \tan 35^{\circ }$

Correct answer:

$72 \tan 35^{\circ }$

Explanation:

The area of a right triangle is half the product of the lengths of its legs, which here are $\overline{AB }$ and $\overline{AC}$ - that is,

$a= \frac{1}{2}(AB)(AC)$

We are given that AC=12 . $\overline{AB }$ is the leg opposite the angle $\angle C$ and $\overline{AC}$ is its adjacent leg, we can find using the tangent ratio:

$\tan (m \angle C) = \frac{AB}{AC}$

Setting AC=12 and $m \angle C = 35 ^{\circ }$ , we get

$\tan 35^{\circ } = \frac{AB}{12}$

Solve for by multiplying both sides by 12:

$12 \cdot \tan 35^{\circ } =12 \cdot \frac{AB}{12}$

$12 \tan 35^{\circ } =AB$

Now, set $AB = 12 \tan 35^{\circ }$ and AC=12 in the area formula:

$a= \frac{1}{2}(AB)(AC)$

$= \frac{1}{2}(12 \tan 35^{\circ })\left (12\right )$

$=72 \tan 35^{\circ }$ ,

the correct choice.

Report an Error

Example Question #1 : Problems Involving Right Triangle Trigonometry

Right

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

Possible Answers:

None of the other choices gives the correct response.

$32 \tan 35^{\circ}$

$64 \tan 35^{\circ}$

$\frac{64}{\tan 35^{\circ}}$

$\frac{32}{\tan 35^{\circ}}$

Correct answer:

$\frac{32}{\tan 35^{\circ}}$

Explanation:

The area of a right triangle is half the product of the lengths of its legs, which here are $\overline{AB }$ and $\overline{AC}$ - that is,

$a= \frac{1}{2}(AB)(AC)$

We are given that AB = 8 . $\overline{AB }$ is the leg opposite the angle $\angle C$ and $\overline{AC}$ is its adjacent leg, we can find using the tangent ratio:

$\tan (m \angle C) = \frac{AB}{AC}$

Setting AB = 8 and $m \angle C = 35 ^{\circ }$ , we get

$\tan 35^{\circ } = \frac{8}{AC}$

Solve for by first, finding the reciprocal of both sides:

$\frac{1}{\tan 35^{\circ }} = \frac{AC}{8}$

Now, multiply both sides by 8:

$\frac{1}{\tan 35^{\circ }} \cdot 8 = \frac{AC}{8} \cdot 8$

$\frac{8}{\tan 35^{\circ }} = AC$

Now, set AB = 8 and $AC = \frac{8}{\tan 35^{\circ }}$ in the area formula:

$a= \frac{1}{2}(AB)(AC)$

$= \frac{1}{2}(8)\left ( \frac{8}{\tan 35^{\circ}} \right )$

$= \frac{1}{2} \left (\frac{8}{1} \right )\left ( \frac{8}{\tan 35^{\circ}} \right )$

$= \frac{32}{\tan 35^{\circ}}$ ,

the correct choice.

Report an Error

Example Question #1 : Problems Involving Right Triangle Trigonometry

$\cos N = \frac{x}{3}$

Evaluate $\tan N$ in terms of .

Possible Answers:

$\frac{\sqrt{x^{2}-9}}{x}$

$\frac{\sqrt{x^{2}-9}}{3}$

$\frac{3}{x}$

$\frac{3}{\sqrt{x^{2}-9}}$

$\frac{x}{\sqrt{x^{2}-9}}$

Correct answer:

$\frac{\sqrt{x^{2}-9}}{x}$

Explanation:

$\cos N = \frac{A}{H} = \frac{x}{3}$

We can set the lengths of the adjacent leg and the hypotenuse to and 3, respectively. By the Pythagorean Theorem, the length of the opposite leg is

$O = \sqrt{H^{2}- A^{2}} = \sqrt{x^{2}-3^{2}} = \sqrt{x^{2}-9 }$

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, so

$\tan N = \frac{O}{A} = \frac{\sqrt{x^{2}-9}}{x}$ .

Report an Error

Example Question #1 : Problems Involving Right Triangle Trigonometry

$\sin a^{\circ } = \frac{4}{7}$

$\tan a^{\circ } = ?$

Possible Answers:

$\frac{\sqrt{33}}{4}$

$\frac{7}{4}$

$\frac{\sqrt{33}}{7}$

$\frac{4}{\sqrt{33}}$

$\frac{7}{\sqrt{33}}$

Correct answer:

$\frac{4}{\sqrt{33}}$

Explanation:

The sine of an angle is defined to be the ratio of the length of the opposite leg of a right triangle to the length of its hypotenuse. Therefore, we can set O = 4, H = 7 . By the Pythagorean Theorem:
the adjacent leg of the triangle has measure

$A = \sqrt{H^{2}-O^{2}} = \sqrt{7^{2}-4^{2}} = \sqrt{49-16} = \sqrt{33}$

The tangent of the angle is the ratio of the length of the opposite leg to that of the adjacent leg, which is

$\tan a ^{\circ }= \frac{O}{A} = \frac{4}{\sqrt{33}}$ ,

the correct response.

Report an Error

View Tutors

Anabel
Certified Tutor

UDELAR, Diploma, Chemistry. IPA, Bachelor of Education, Mathematics.

View Tutors

Ruman
Certified Tutor

MBA, Master's/Graduate, Human resources .

View Tutors

Himanshi
Certified Tutor

teerthanker mahaveer University , Master's/Graduate, Mba.

All HiSET: Math Resources

125 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ACT Tutors in Washington DC, ACT Tutors in Los Angeles, ISEE Tutors in Phoenix, Physics Tutors in Philadelphia, MCAT Tutors in Philadelphia, French Tutors in Denver, Biology Tutors in San Diego, Reading Tutors in New York City, MCAT Tutors in Boston, English Tutors in Atlanta

Popular Courses & Classes

MCAT Courses & Classes in Miami, LSAT Courses & Classes in San Diego, SAT Courses & Classes in Miami, SAT Courses & Classes in San Diego, GMAT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Atlanta, MCAT Courses & Classes in Houston, LSAT Courses & Classes in Houston, SAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Atlanta

Popular Test Prep

MCAT Test Prep in Phoenix, ACT Test Prep in Boston, SAT Test Prep in Miami, ISEE Test Prep in San Francisco-Bay Area, MCAT Test Prep in New York City, SAT Test Prep in Boston, ISEE Test Prep in Atlanta, GMAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Denver, SAT Test Prep in Chicago

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

HiSET: Math : Problems involving right triangle trigonometry

Study concepts, example questions & explanations for HiSET: Math

All HiSET: Math Resources

Example Questions

Example Question #1 : Sine

Example Question #1 : Sine

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Problems Involving Right Triangle Trigonometry

Example Question #1 : Problems Involving Right Triangle Trigonometry

All HiSET: Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: