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Math
GeometryElementary Geometry for College Students6th Edition

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Chapter 11.2, Problem 41E
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Elementary Geometry for College St...

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

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Chapter
Section
Problem 1E
Problem 2E
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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
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Problem 29E
Problem 30E
Problem 31E
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Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
BuyFindarrow_forward

Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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Use the drawing provided to show that the area of the isosceles triangle is A = s 2 sin θ cos θ

Chapter 11.2, Problem 41E, Use the drawing provided to show that the area of the isosceles triangle is A=s2sincos

To determine

To find:

To show that the area of the isosceles triangle is A=s2sinθcosθ by using the following figure,

Elementary Geometry for College Students, Chapter 11.2, Problem 41E

Explanation

Consider the following figure,

Let us put the name for the isosceles triangle ia ABC and altitude is AD¯

We know that BD=DC=base2

Assume that BD=DC=x units

General formula for cosine ratio and sin ratio is given below,

cosθ=adjacenthypotenuseθ=cos−1(adjacenthypotenuse) and sinθ=oppositehypotenuseθ=sin−1(oppositehypotenuse)

tanθ=oppositeadjacent

For ΔADB:

From the above figure shows that the s is the value of hypotenuse since it is opposite to right angle.

Then DC is the length of the leg opposite to θ.

Therefore,

cosθ=ADAB=ADsAD=(cosθ)s

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