/*! 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 21 Let \(A B C\) be a triangle. We ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(A B C\) be a triangle. We use the standard labelling convention, whereby the side \(B C\) opposite \(A\) has length \(a\), the side \(C A\) opposite \(B\) has length \(b,\) and the side \(A B\) opposite \(C\) has length \(c\). Prove that, if \(c^{2}=a^{2}+b^{2},\) then \(\angle B C A\) is a right angle.

Short Answer

Expert verified
If \(c^2 = a^2 + b^2\), then \(\angle BCA\) is a right angle according to the Pythagorean Theorem.

Step by step solution

01

Understand the Problem

We have a triangle with sides labeled as follows: side \(BC\) opposite angle \(A\) is \(a\), side \(CA\) opposite angle \(B\) is \(b\), and side \(AB\) opposite angle \(C\) is \(c\). We're given that \(c^2 = a^2 + b^2\) and need to prove that \(\angle BCA\) is a right angle.
02

Recall the Pythagorean Theorem

The Pythagorean Theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Therefore, if \(c^2 = a^2 + b^2\), triangle \(ABC\) with \(c\) as the hypotenuse implies that the triangle is right-angled opposite to \(c\).
03

Identify the Triangle Configuration

In triangle \(ABC\), \(c\) is opposite to angle \(C\). If \(c^2 = a^2 + b^2\), then the configuration states that the side opposite the right angle \(\angle BCA\) is \(c\), meaning \(\angle BCA\) should be the right angle.
04

Conclude the Proof

Given the conditions of the Pythagorean Theorem that have been satisfied (\(c^2 = a^2 + b^2\)), and the role of the sides within the theorem, it follows necessarily that \(\angle BCA\) is a right angle.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Geometry
Geometry is the mathematical study concerned with the properties and relations of points, lines, surfaces, and solids. It involves shapes and figures and how these can be placed, move, and interact within space. At its core, geometry helps us understand the physical world by providing a framework to analyze the shapes we see.
In geometry, we often deal with different types of figures like angles, circles, and triangles. Each of these figures has specific properties and rules they follow. Using these properties, we solve various problems and make deductions, just like proving if a triangle is right-angled using the Pythagorean Theorem.
Understanding geometry lays the foundation for more advanced topics in mathematics. It is fundamental in fields ranging from art and architecture to physics and engineering.
  • It helps us calculate dimensions, areas, and volumes.
  • Provides tools to describe locations in space.
  • Helps us prove mathematical relations like the Pythagorean Theorem.
The Special Right Triangle
A right triangle is a special type of triangle where one of its angles is exactly 90 degrees. It is one of the most important geometric figures in mathematics due to its unique properties and the ease with which it allows mathematical operations.
In a right triangle, the side opposite the right angle is known as the hypotenuse. It is always the longest side of the triangle. The other two sides are referred to as the legs of the triangle.
Right triangles not only help in verifying the Pythagorean Theorem but are also central to trigonometry, which deals with the relationships between the angles and sides of triangles.
In practical applications, right triangles are everywhere. From construction to navigation, this simple shape helps solve complex problems. Understanding its properties allows us to utilize it in various scenarios effectively.
  • The hypotenuse is opposite the right angle.
  • Right triangles simplify calculations involving height and distance.
  • They underpin various mathematical theorems and principles.
Exploring Triangle Properties
Triangles come with a rich set of properties that not only make them interesting but also highly useful in mathematics and real-life applications.
Each triangle has three sides and three angles, leading to several types like equilateral, isosceles, and scalene triangles. However, when we talk about the Pythagorean Theorem, we are specifically interested in right triangles.
One key property of triangles is that the sum of their internal angles is always 180 degrees. This property helps in solving many problems, especially those involving unknown angles. In a right triangle, since one angle is always 90 degrees, the sum of the other two angles must be 90 degrees.
  • Triangles can be categorized based on side length and angle measurement.
  • Understanding angle sums helps with angle calculation.
  • Properties help us apply theorems like the Pythagorean Theorem.
These properties are applied when using concepts like the Pythagorean Theorem, which ties together the lengths of the sides in a right triangle, providing powerful methods for proving various geometric truths.

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

The 4 by 4 "multiplication table" below is completely familiar. \(\begin{array}{rrrr}1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \\ 4 & 8 & 12 & 16\end{array}\) What is the total of all the numbers in the 4 by 4 square? How should one write this answer in a way that makes the total obvious?

(a) Joining the midpoints of the edges of an equilateral triangle \(A B C\) cuts the triangle into four identical smaller equilateral triangles. Removing one of the three outer small triangles (say \(A M N\), with \(M\) on \(A C\) ) leaves three-quarters of the original shape in the form of an isosceles trapezium \(M N B C .\) Show how to cut this isosceles trapezium into four congruent pieces. (b) Joining the midpoints of opposite sides of a square cuts the square into four congruent smaller squares. If we remove one of these squares, we are left with three-quarters of the original square in the form of an L-shape. Show how to cut this L-shape into four congruent pieces.

(Pages of a newspaper) I found a (double) sheet from an old newspaper, with pages 14 and 27 next to each other. How many pages were there in the original newspaper?

(a) Use Pythagoras' Theorem in a square \(A B C D\) of side 1 to show that the diagonal \(A C\) has length \(\sqrt{2}\). Use this to work out in your head the exact values of \(\sin 45^{\circ}, \cos 45^{\circ}, \tan 45^{\circ}\) (b) In an equilateral triangle \(\triangle A B C\) with sides of length \(2,\) join \(A\) to the midpoint \(M\) of the base \(B C\). Apply Pythagoras' Theorem to find \(A M\). Hence work out in your head the exact values of \(\sin 30^{\circ}, \cos 30^{\circ}, \tan 30^{\circ},\) \(\sin 60^{\circ}, \cos 60^{\circ}, \tan 60^{\circ}\) (c)(i) On the unit circle with centre at the origin \(O:(0,0),\) mark the point \(P\) so that \(P\) lies in the first quadrant, and so that \(O P\) makes an angle \(\theta\) with the positive \(x\) -axis (measured anticlockwise from the positive \(x\) -axis). Explain why \(P\) has coordinates \((\cos \theta, \sin \theta)\). (ii) Extend the definitions of \(\cos \theta\) and \(\sin \theta\) to apply to angles beyond the first quadrant, so that for any point \(P\) on the unit circle, where \(O P\) makes an angle \(\theta\) measured anticlockwise from the positive \(x\) -axis, the coordinates of \(P\) are \((\cos \theta, \sin \theta) .\) Check that the resulting functions sin and cos satisfy: \(*\) sin and cos are both positive in the first quadrant, \(*\) sin is positive and cos is negative in the second quadrant, \(*\) sin and cos are both negative in the third quadrant, and \(*\) sin is negative and \(\cos\) is positive in the fourth quadrant. (iii) Use (a), (b) to calculate the exact values of \(\cos 315^{\circ}, \sin 225^{\circ}, \tan 210^{\circ}\), \(\cos 120^{\circ}, \sin 960^{\circ}, \tan \left(-135^{\circ}\right)\) (d) Given a circle of radius \(1,\) work out the exact area of a regular \(n\) -gon inscribed in the circle: (i) when \(n=3\) (ii) when \(n=4\) (iii) when \(n=6\) (iv) when \(n=8\) (v) when \(n=12\). (e) Given a circle of radius 1 , work out the area of a regular \(n\) -gon circumscribed around the circle: (i) when \(n=3\) (ii) when \(n=4\) (iii) when \(n=6\) (iv) when \(n=8\) (v) when \(n=12\). Knowing the exact values of \(\sin , \cos\) and tan for the special angles \(0^{\circ}\), \(30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}\) is like knowing one's tables. In particular, it allows one to evaluate trigonometric functions mentally for related angles in all four quadrants (using the CAST mnemonic \(-\mathrm{C}\) being in the \(\mathrm{SE}\) of the unit circle, \(A\) in the NE quadrant, \(S\) in the NW quadrant, and \(T\) in the SW quadrant - to remind us which functions are positive in each quadrant). These special angles arise over and over again in connection with equilateral triangles, squares, regular hexagons, regular octagons, regular dodecagons, etc., where one can use what one knows to calculate exactly in geometry.

Problem 39 (Shadows) Can one use the Sun's rays to produce a plane shadow of a cube: (i) in the form of an equilateral triangle? (ii) in the form of a square? (iii) in the form of a pentagon? (iv) in the form of a regular hexagon? (v) in the form of a polygon with more than six sides?

See all solutions

Recommended explanations on Math Textbooks

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.