A right triangle for the given angle and the sides of the triangle are to be labelled as positive or negative. Label the sides of the triangle as positive or negative. To find: The reference angles and functions of the angles. The value of sine, cosine, tangent, cotangent, cosecant, secant for a given angle, up to 4 decimals.

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Answer 1

Question

Chapter 71, Problem 13A

To determine

To sketch:

A right triangle for the given angle and the sides of the triangle are to be labelled as positive or negative.

Label the sides of the triangle as positive or negative.

To find: The reference angles and functions of the angles.

The value of sine, cosine, tangent, cotangent, cosecant, secant for a given angle, up to 4 decimals.

Expert Solution & Answer

Answer to Problem 13A

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Explanation of Solution

Given Information:

θ= 208°50'

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 1

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 2

In Quadrant 3, y is negative and x is negative.

In the above triangle, r is the hypotenuse, the adjacent side x is negative, the opposite side y is negative.

Reference angle∠α = 208°50’−180°∠α =28.83°

sinα=−yrIt is a negative function.sin208°50’=sin(208°50’−180°)sin208°50’=−sin28.83°sin208°50’=−0.4822cosα=−xrIt is a negative function.cos208°50’=cos(208°50’−180°)cos208°50’=−cos28.83°cos208°50’=−0.876tanα=−y−xIt is a positive functiontan208°50’=tan(208°50’−180°)tan208°50’=tan28.83°tan208°50’=0.5504cotα=−x−yIt is a positive functioncot208°50’=cot(208°50’−180°)cot208°50’=−cot28.83°cot208°50’=1.8167secα=r−xIt is a negative function.sec208°50’=sec(208°50’−180°)sec208°50’=sec28.83°sec208°50’=−1.1414cosec α=r−yIt is a negative function.cosec208°50’=cosec(208°50’−180°)cosec208°50’=−cosec28.83°cosec208°50’=−2.0737

Conclusion:

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

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Answer 2

Question

Chapter 71, Problem 13A

To determine

To sketch:

A right triangle for the given angle and the sides of the triangle are to be labelled as positive or negative.

Label the sides of the triangle as positive or negative.

To find: The reference angles and functions of the angles.

The value of sine, cosine, tangent, cotangent, cosecant, secant for a given angle, up to 4 decimals.

Expert Solution & Answer

Answer to Problem 13A

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Explanation of Solution

Given Information:

θ= 208°50'

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 1

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 2

In Quadrant 3, y is negative and x is negative.

In the above triangle, r is the hypotenuse, the adjacent side x is negative, the opposite side y is negative.

Reference angle∠α = 208°50’−180°∠α =28.83°

sinα=−yrIt is a negative function.sin208°50’=sin(208°50’−180°)sin208°50’=−sin28.83°sin208°50’=−0.4822cosα=−xrIt is a negative function.cos208°50’=cos(208°50’−180°)cos208°50’=−cos28.83°cos208°50’=−0.876tanα=−y−xIt is a positive functiontan208°50’=tan(208°50’−180°)tan208°50’=tan28.83°tan208°50’=0.5504cotα=−x−yIt is a positive functioncot208°50’=cot(208°50’−180°)cot208°50’=−cot28.83°cot208°50’=1.8167secα=r−xIt is a negative function.sec208°50’=sec(208°50’−180°)sec208°50’=sec28.83°sec208°50’=−1.1414cosec α=r−yIt is a negative function.cosec208°50’=cosec(208°50’−180°)cosec208°50’=−cosec28.83°cosec208°50’=−2.0737

Conclusion:

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

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Answer 3

Question

Chapter 71, Problem 13A

To determine

To sketch:

A right triangle for the given angle and the sides of the triangle are to be labelled as positive or negative.

Label the sides of the triangle as positive or negative.

To find: The reference angles and functions of the angles.

The value of sine, cosine, tangent, cotangent, cosecant, secant for a given angle, up to 4 decimals.

Expert Solution & Answer

Answer to Problem 13A

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Explanation of Solution

Given Information:

θ= 208°50'

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 1

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 2

In Quadrant 3, y is negative and x is negative.

In the above triangle, r is the hypotenuse, the adjacent side x is negative, the opposite side y is negative.

Reference angle∠α = 208°50’−180°∠α =28.83°

sinα=−yrIt is a negative function.sin208°50’=sin(208°50’−180°)sin208°50’=−sin28.83°sin208°50’=−0.4822cosα=−xrIt is a negative function.cos208°50’=cos(208°50’−180°)cos208°50’=−cos28.83°cos208°50’=−0.876tanα=−y−xIt is a positive functiontan208°50’=tan(208°50’−180°)tan208°50’=tan28.83°tan208°50’=0.5504cotα=−x−yIt is a positive functioncot208°50’=cot(208°50’−180°)cot208°50’=−cot28.83°cot208°50’=1.8167secα=r−xIt is a negative function.sec208°50’=sec(208°50’−180°)sec208°50’=sec28.83°sec208°50’=−1.1414cosec α=r−yIt is a negative function.cosec208°50’=cosec(208°50’−180°)cosec208°50’=−cosec28.83°cosec208°50’=−2.0737

Conclusion:

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

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Answer 4

Question

Chapter 71, Problem 13A

To determine

To sketch:

A right triangle for the given angle and the sides of the triangle are to be labelled as positive or negative.

Label the sides of the triangle as positive or negative.

To find: The reference angles and functions of the angles.

The value of sine, cosine, tangent, cotangent, cosecant, secant for a given angle, up to 4 decimals.

Expert Solution & Answer

Answer to Problem 13A

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Explanation of Solution

Given Information:

θ= 208°50'

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 1

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 2

In Quadrant 3, y is negative and x is negative.

In the above triangle, r is the hypotenuse, the adjacent side x is negative, the opposite side y is negative.

Reference angle∠α = 208°50’−180°∠α =28.83°

sinα=−yrIt is a negative function.sin208°50’=sin(208°50’−180°)sin208°50’=−sin28.83°sin208°50’=−0.4822cosα=−xrIt is a negative function.cos208°50’=cos(208°50’−180°)cos208°50’=−cos28.83°cos208°50’=−0.876tanα=−y−xIt is a positive functiontan208°50’=tan(208°50’−180°)tan208°50’=tan28.83°tan208°50’=0.5504cotα=−x−yIt is a positive functioncot208°50’=cot(208°50’−180°)cot208°50’=−cot28.83°cot208°50’=1.8167secα=r−xIt is a negative function.sec208°50’=sec(208°50’−180°)sec208°50’=sec28.83°sec208°50’=−1.1414cosec α=r−yIt is a negative function.cosec208°50’=cosec(208°50’−180°)cosec208°50’=−cosec28.83°cosec208°50’=−2.0737

Conclusion:

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

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Answer 5

Chapter 71, Problem 13A

To determine

To sketch:

A right triangle for the given angle and the sides of the triangle are to be labelled as positive or negative.

Label the sides of the triangle as positive or negative.

To find: The reference angles and functions of the angles.

The value of sine, cosine, tangent, cotangent, cosecant, secant for a given angle, up to 4 decimals.

Expert Solution & Answer

Answer to Problem 13A

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Explanation of Solution

Given Information:

θ= 208°50'

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 1

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 2

In Quadrant 3, y is negative and x is negative.

In the above triangle, r is the hypotenuse, the adjacent side x is negative, the opposite side y is negative.

Reference angle∠α = 208°50’−180°∠α =28.83°

sinα=−yrIt is a negative function.sin208°50’=sin(208°50’−180°)sin208°50’=−sin28.83°sin208°50’=−0.4822cosα=−xrIt is a negative function.cos208°50’=cos(208°50’−180°)cos208°50’=−cos28.83°cos208°50’=−0.876tanα=−y−xIt is a positive functiontan208°50’=tan(208°50’−180°)tan208°50’=tan28.83°tan208°50’=0.5504cotα=−x−yIt is a positive functioncot208°50’=cot(208°50’−180°)cot208°50’=−cot28.83°cot208°50’=1.8167secα=r−xIt is a negative function.sec208°50’=sec(208°50’−180°)sec208°50’=sec28.83°sec208°50’=−1.1414cosec α=r−yIt is a negative function.cosec208°50’=cosec(208°50’−180°)cosec208°50’=−cosec28.83°cosec208°50’=−2.0737

Conclusion:

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Answer 6

Chapter 71, Problem 13A

To determine

To sketch:

A right triangle for the given angle and the sides of the triangle are to be labelled as positive or negative.

Label the sides of the triangle as positive or negative.

To find: The reference angles and functions of the angles.

The value of sine, cosine, tangent, cotangent, cosecant, secant for a given angle, up to 4 decimals.

Expert Solution & Answer

Answer to Problem 13A

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Explanation of Solution

Given Information:

θ= 208°50'

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 1

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 2

In Quadrant 3, y is negative and x is negative.

In the above triangle, r is the hypotenuse, the adjacent side x is negative, the opposite side y is negative.

Reference angle∠α = 208°50’−180°∠α =28.83°

sinα=−yrIt is a negative function.sin208°50’=sin(208°50’−180°)sin208°50’=−sin28.83°sin208°50’=−0.4822cosα=−xrIt is a negative function.cos208°50’=cos(208°50’−180°)cos208°50’=−cos28.83°cos208°50’=−0.876tanα=−y−xIt is a positive functiontan208°50’=tan(208°50’−180°)tan208°50’=tan28.83°tan208°50’=0.5504cotα=−x−yIt is a positive functioncot208°50’=cot(208°50’−180°)cot208°50’=−cot28.83°cot208°50’=1.8167secα=r−xIt is a negative function.sec208°50’=sec(208°50’−180°)sec208°50’=sec28.83°sec208°50’=−1.1414cosec α=r−yIt is a negative function.cosec208°50’=cosec(208°50’−180°)cosec208°50’=−cosec28.83°cosec208°50’=−2.0737

Conclusion:

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Answer 7

Chapter 71, Problem 13A

To determine

To sketch:

A right triangle for the given angle and the sides of the triangle are to be labelled as positive or negative.

Label the sides of the triangle as positive or negative.

To find: The reference angles and functions of the angles.

The value of sine, cosine, tangent, cotangent, cosecant, secant for a given angle, up to 4 decimals.

Answer 8

Expert Solution & Answer

Answer to Problem 13A

Reference angle is 28°50'.

Sin 208°50' is -0.4822. It is a negative function.

Cos 208°50' is -0.876. It is a negative function.

Tan 208°50' is 0.5504. It is a positive function.

Cot 208°50' is 1.8167. It is a positive function.

Sec 208°50' is -1.1414. It is a negative function.

Cosec 208°50' is -2.073. It is a negative function.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given Information:

θ= 208°50'

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 1

Mathematics For Machine Technology, Chapter 71, Problem 13A , additional homework tip 2

In Quadrant 3, y is negative and x is negative.

In the above triangle, r is the hypotenuse, the adjacent side x is negative, the opposite side y is negative.

Reference angle∠α = 208°50’−180°∠α =28.83°

sinα=−yrIt is a negative function.sin208°50’=sin(208°50’−180°)sin208°50’=−sin28.83°sin208°50’=−0.4822cosα=−xrIt is a negative function.cos208°50’=cos(208°50’−180°)cos208°50’=−cos28.83°cos208°50’=−0.876tanα=−y−xIt is a positive functiontan208°50’=tan(208°50’−180°)tan208°50’=tan28.83°tan208°50’=0.5504cotα=−x−yIt is a positive functioncot208°50’=cot(208°50’−180°)cot208°50’=−cot28.83°cot208°50’=1.8167secα=r−xIt is a negative function.sec208°50’=sec(208°50’−180°)sec208°50’=sec28.83°sec208°50’=−1.1414cosec α=r−yIt is a negative function.cosec208°50’=cosec(208°50’−180°)cosec208°50’=−cosec28.83°cosec208°50’=−2.0737