3rd Edition

ISBN: 9781285607160

Author: Larson

Publisher: CENGAGE CO

Concept explainers

Trigonometry in mathematics deals with the right-angled triangle’s angles and sides. By trigonometric identities, we mean the identities we use whenever we need to express the various trigonometric functions in terms of an equation.

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse of normal trigonometric functions. Alternatively denoted as cyclometric or arcus functions, these inverse trigonometric functions exist to counter the basic trigonometric functions, such as sine (sin), cosi…

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Question

Chapter 5.2, Problem 66E

To determine

To verify: s=hcotθ

Expert Solution

Explanation of Solution

Given:The length s of a shadow cast by a vertical gnomom (a device used to tell time) of height h when the angle of the sun above the horizon is θ can be modeled by the equation s=hsin( 90∘−θ)sinθ .

Calculation:

The length of shadow cast by a vertical gnomon, s=hsin( 90∘−θ)sinθ

s=hcosθsinθ Cofunction Identitys=hcotθ Quotient Identity

Hence verified

To determine

To complete: s=hcotθ , the table at different value of θ.

Expert Solution

Answer to Problem 66E

θ 15∘ 30∘ 45∘ 60∘ 75∘ 90∘s18.668.6652.891.340

Explanation of Solution

Calculation:

The length of shadow cast by a vertical gnomon, s=hcotθ

For h=5 feet , make the graph of s=5cotθ .

EBK PRECALCULUS W/LIMITS, Chapter 5.2, Problem 66E

Using graph to complete the table.

θ 15∘ 30∘ 45∘ 60∘ 75∘ 90∘s18.668.6652.891.340

To determine

To find:The angle of the sun that result in the maximum and minimum length of the shadow.

Expert Solution

Answer to Problem 66E

Maximum at 15∘

Minimum at 90∘

Explanation of Solution

Given:The given table,

θ 15∘ 30∘ 45∘ 60∘ 75∘ 90∘s18.668.6652.891.340

Calculation:

The length of shadow cast by a vertical gnomon, s=5cotθ

From table,

The maximum value of smax=18.66 at θ=15∘

The minimum value of smin=0 at θ=90∘

To determine

To find:The time when the angle of the sun above the horizon is 90∘.

Expert Solution

Answer to Problem 66E

12:00 PM

Explanation of Solution

Given:The given table,

θ 15∘ 30∘ 45∘ 60∘ 75∘ 90∘s18.668.6652.891.340

Calculation:

The length of shadow cast by a vertical gnomon, s=5cotθ

From table,

The maximum value of smax=18.66 at θ=15∘

The minimum value of smin=0 at θ=90∘

The length of shadow is 0 just above 90∘ of horizon.

The time will be 12:00 PM

Chapter 5.2, Problem 65E

Chapter 5.2, Problem 67E

Chapter 5 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E

Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - Prob. 49E Ch. 5.1 - Prob. 50E Ch. 5.1 - Prob. 51E Ch. 5.1 - Prob. 52E Ch. 5.1 - Prob. 53E Ch. 5.1 - Prob. 54E Ch. 5.1 - Prob. 55E Ch. 5.1 - Prob. 56E Ch. 5.1 - Prob. 57E Ch. 5.1 - Prob. 58E Ch. 5.1 - Prob. 59E Ch. 5.1 - Prob. 60E Ch. 5.1 - 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