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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 11.1, Problem 36E
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Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

Solutions

Chapter
Section
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
BuyFindarrow_forward

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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For Exercises 35 to 38, make drawings as needed.

In right △ A B C , γ = 90 ° and β = 55 ° . If AB=20in., find

a) a (the length of B C - ) correct to tenths.

b) b (the length of A C - ) correct to tenths.

c) the area of right △ A B C .

To determine

To find:

a. The length of a.

Explanation

In a right angle triangle,

1) The side PQ, which is opposite to the right angle PRQ is called the hypotenuse.

(The hypotenuse is the longest side of the right triangle.)

2) The side RQ is called the adjacent side of angle θ.

3) The side PR is called the opposite side of angle θ.

Hence, the sine ratio for an acute angle is oppositehypotenuse

Given:

In a right angled triangle ABC,

AB = 20 in, β=55°,γ=90°

Hypotenuse, AC=b.

Let AB be the opposite side of the triangle.

θ=55°

We know that,

sinθ=oppositeHypotenuse

sin 55°=20b

From the table 11

To determine

To find:

b. The length of b.

To determine

To find:

c. The area of the triangle, ABC.

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Chapter 11.1,
Problem 35E
Chapter 11.1,
Problem 37E
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Sect-11.1 P-1ESect-11.1 P-2ESect-11.1 P-3ESect-11.1 P-4ESect-11.1 P-5ESect-11.1 P-6ESect-11.1 P-7ESect-11.1 P-8ESect-11.1 P-9ESect-11.1 P-10E
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