/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Use the previous problem to show... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 35

Use the previous problem to show that in every triangle, the sum of the lengths of any two sides is greater than the length of the third side.

Short Answer

Expert verified
Using the angles within a triangle, we can prove the Triangle Inequality Theorem that states the sum of the lengths of any two sides is greater than the length of the third side. We can show that: 1. If angle A and angle B add up to less than 180掳, then a + b > c 2. If angle B and angle C add up to less than 180掳, then b + c > a 3. If angle A and angle C add up to less than 180掳, then c + a > b Hence, this proves the Triangle Inequality Theorem for any triangle with sides a, b, and c.

Step by step solution

01

Identify the sides of the triangle

To start with, let's name the sides of the triangle as a, b, and c. We will prove that the sum of the lengths of any two sides of the triangle is greater than the length of the third side.
02

Apply the previous problem's information

Using the information from the previous problem, we know that if angle A is opposite to side a, angle B is opposite to side b, and angle C is opposite to side c, the angles within the triangle must satisfy the following relationships: angle A + angle B + angle C = 180掳
03

Prove that a + b > c

Let's start by proving that a + b > c. We know that angle C is opposite to side c, so for the sum of angles A and B, we have: angle A + angle B = 180掳 - angle C Since angles A and B add up to less than 180掳, the sum of their corresponding sides must be greater than side c. Symbolically, we can write this as: a + b > c
04

Prove that b + c > a

Similarly, let's prove that b + c > a. Since angle A is opposite to side a, for the sum of angles B and C, we have: angle B + angle C = 180掳 - angle A Again, since angles B and C add up to less than 180掳, the sum of their corresponding sides must be greater than side a. Symbolically, we can write this as: b + c > a
05

Prove that c + a > b

Lastly, let's prove that c + a > b. Since angle B is opposite to side b, for the sum of angles A and C, we have: angle A + angle C = 180掳 - angle B Yet again, since angles A and C add up to less than 180掳, the sum of their corresponding sides must be greater than side b. Symbolically, we can write this as: c + a > b
06

Conclusion

From our proofs in steps 3, 4, and 5, we have shown that for any triangle with sides a, b, and c: a + b > c b + c > a c + a > b And with these relationships established, we have successfully proven that the sum of the lengths of any two sides of a triangle is always greater than the length of the third side, demonstrating the Triangle Inequality Theorem.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Evaluate the indicated expressions assuming that $$ \begin{array}{ll} \cos x=\frac{1}{3} & \text { and } \sin y=\frac{1}{4} \\ \sin u=\frac{2}{3} & \text { and } \cos v=\frac{1}{5} \end{array} $$ Assume also that \(x\) and \(u\) are in the interval \(\left(0, \frac{\pi}{2}\right),\) that \(y\) is in the interval \(\left(\frac{\pi}{2}, \pi\right),\) and that \(v\) is in the interval \(\left(-\frac{\pi}{2}, 0\right) .\) $$\cos (u-v)$$

Show that $$\cos x-\cos y=2 \sin \frac{x+y}{2} \sin \frac{y-x}{2}$$ for all \(x, y\).

Find an exact expression for \(\sin 15^{\circ}\).

Find an exact expression for \(\sin \frac{\pi}{24}\).

Show that \(\cos 20^{\circ}\) is a zero of the polynomial \(8 x^{3}-6 x-1\) [Hint: Set \(\theta=20^{\circ}\) in the identity from the previous problem.]
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.