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Chapter 1: Problem 33

Let \(\Delta=\operatorname{area}(\triangle A B C)\) (a) Prove that $$ \Delta=\frac{1}{2} \cdot a b \cdot \sin C . $$ (b) Prove that \(4 R \Delta=a b c\).

Short Answer

Expert verified
(a) Use formula: \( \Delta = \frac{1}{2}ab \sin C \). (b) Apply law of sines to show \( 4R \Delta = abc \).

Step by step solution

01

Calculate Area Using Trigonometry

The formula for the area of a triangle using two sides and the sine of the included angle is given by the expression \( \Delta = \frac{1}{2} \cdot a \cdot b \cdot \sin C \) . To derive this, recall the broader trigonometric identity for calculating area: if C is the angle between sides a and b of a triangle, the area is calculated as half the product of the lengths of these sides multiplied by the sine of the angle between them. This relationship arises from the definition of sine in the context of right triangles and is extended here to any triangle through the law of sines.
02

Use Law of Sines for Circumradius

The law of sines states that for any triangle, \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \), where \( R \) is the circumradius. To solve part (b), you first apply the law of sines to express the sides in terms of angles and the circumradius. From this, rearrange it to solve for a, b, and c involving \( R \).
03

Calculate Relationship for Side-Product

Using the circumradius relation: \( a = 2R \sin A \), \( b = 2R \sin B \), and \( c = 2R \sin C \), substitute these into a product form: \( abc = (2R \sin A)(2R \sin B)(2R \sin C) = 8R^3 \sin A \sin B \sin C \).
04

Establish the Equality

For the expression \( 4R\Delta \), substitute \( \Delta = \frac{1}{2}ab\sin C\) derived from step 1: \( 4R \cdot \frac{1}{2} \cdot ab \cdot \sin C = 2R \cdot ab \cdot \sin C \). Now using the identities from step 3 substitute back the sine of angles: \( 2R \cdot 2R \sin A \cdot 2R \sin B \cdot \sin C = abc \). This verifies that the expressions \( 4R\Delta \) and \( abc \) are indeed equivalent.

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Key Concepts

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

Area of Triangle
Finding the area of a triangle is possible with several methods, but using trigonometry provides a powerful and versatile approach. The formula \( \Delta = \frac{1}{2} \cdot a \cdot b \cdot \sin C \) allows us to determine the area of a triangle when we know two sides and the included angle between them.
This method is particularly useful when dealing with non-right triangles where traditional base-height methods may not be applicable.
  • The formula comes from the trigonometric definition of the sine function in angles, which represents the ratio of the opposite side to the hypotenuse in a right-angled triangle.
  • It considers the angle \( C \) as creating a "right angle" abstraction, allowing the otherwise complicated geometry to be easily computed.
  • This formula is elegantly derived and connected closely to the wider trigonometric concepts that extend beyond just right triangles.
Understanding this formula strengthens your grasp of triangles in geometry, providing foundational knowledge that can be applied as problems increase in complexity.
Law of Sines
The Law of Sines is a revealing theorem that serves as a bridge between side lengths and angles in any triangle. Expressed as \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \), it establishes an important relationship in terms of the circumradius \( R \).
This makes it an essential tool for solving triangles when certain information is missing.
  • The law shows that the ratio of a side length to the sine of its opposite angle is constant for all sides of a triangle, where this constant is connected to the circumradius \( R \).
  • It is often employed in scenarios where we know two angles and one side (AAS or ASA situations), or two sides and the angle opposite one of them (SSA situation).
  • It complements the Law of Cosines, and both are extensions for analyzing non-right triangles.
The power of the Law of Sines lies in its ability to simplify and solve for unknowns in geometric problems, making it an invaluable part of your mathematical toolkit.
Circumradius
The circumradius, denoted as \( R \), is a fundamental concept when examining the properties of a triangle's circumcircle - the circle that passes through all three vertices. An understanding of this radius ties together many geometric properties, including the Law of Sines, and helps in proving several crucial relationships involving a triangle's area.
  • The circumradius is calculated using the formula \( R = \frac{abc}{4\Delta} \), which shows its direct relationship to the side lengths \( a, b, \) and \( c \), as well as the area \( \Delta \).
  • It allows us to develop expressions that show equivalences, such as \( 4R\Delta = abc \), which are used in various proofs and derivations, as seen in our problem.
  • The concept illustrates the symmetrical properties of triangles, with \( R \) serving as a universal measure linking diverse triangular properties.
Mastering the circumradius concept will aid in understanding complex geometrical relationships and provide an insightful view into how sides and angles interact within triangles.

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Most popular questions from this chapter

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?

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.

(a) Expand and simplify in your head: (i) \((\sqrt{2}+1)^{2}\) (ii) \((\sqrt{2}-1)^{2}\) (iii) \((1+\sqrt{2})^{3}\) (b) Simplify: (i) \(\sqrt{10+4 \sqrt{6}}\) (ii) \(\sqrt{5+2 \sqrt{6}}\) (iii) \(\sqrt{\frac{3+\sqrt{5}}{2}}\) (iv) \(\sqrt{10-2 \sqrt{5}}\) The expressions which occur in exercises to develop fluency in working with surds often appear arbitrary. But they may not be. The arithmetic of surds arises naturally: for example, some of the expressions in the previous problem have already featured in Problem \(\mathbf{3}(\mathrm{c}) .\) In particular, surds will feature whenever Pythagoras' Theorem is used to calculate lengths in geometry, or when a proportion arising from similar triangles requires us to solve a quadratic equation. So surd arithmetic is important. For example: \- A regular octagon with side length 1 can be surrounded by a square of side \(\sqrt{2}+1\) (which is also the diameter of its incircle); so the area of the regular octagon equals \((\sqrt{2}+1)^{2}-1\) (the square minus the four corners). \- \(\sqrt{2}-1\) features repeatedly in the attempt to apply the Euclidean algorithm, or anthyphairesis, to express \(\sqrt{2}\) as a "continued fraction". -\(\sqrt{10-2 \sqrt{5}}\) may look like an arbitrary, uninteresting repeated surd, but is in fact very interesting, and has already featured as \(4 \sin 36^{\circ}\) in Problem \(\mathbf{3}(\mathrm{c})\) \- One of the simplest ruler and compasses constructions for a regular pentagon \(A B C D E\) (see Problem 185) starts with a circle of radius 2 , centre \(O,\) and a point \(A\) on the circle, and in three steps constructs the next point \(B\) on the circle, where \(\underline{A B}\) is an edge of the inscribed regular pentagon, and $$ \underline{A B}=\sqrt{10-2 \sqrt{5}} $$

(Pythagoras' Theorem) Let \(\triangle A B C\) be a right angled triangle, with a right angle at \(C .\) Draw the squares \(A C Q P, C B S R,\) and \(B A U T\) on the three sides, external to \(\triangle A B C\). Use the resulting diagram to prove in your head that the square \(B A U T\) on \(B A\) is equal to the sum of the other two squares by: \- drawing the line through \(C\) perpendicular to \(A B\), to meet \(A B\) at \(X\) and \(U T\) at \(Y\) \- observing that \(P A\) is parallel to \(Q C B\), so that \(\triangle A C P\) (half of the square \(A C Q P,\) with base \(A P\) and perpendicular height \(A C)\) is equal in area to \(\triangle A B P\) (with base \(A P\) and the same perpendicular height) \- noting that \(\triangle A B P\) is SAS-congruent to \(\triangle A U C,\) and that \(\triangle A U C\) is equal in area to \(\triangle A U X\) (half of rectangle \(A U Y X,\) with base \(A U\) and height \(A X)\) \- whence \(A C Q P\) is equal in area to rectangle \(A U Y X\) \- similarly \(B C R S\) is equal in area to \(B T Y X\). The proof in Problem 18 is the proof to be found in Euclid's Elements Book 1, Proposition 47. Unlike many proofs, \- it is clear what the proof depends on (namely SAS triangle congruence, and the area of a triangle), and \- it reveals exactly how the square on the hypotenuse \(A B\) divides into two summands \(-\) one equal to the square on \(A C\) and one equal to the square on \(B C\).

(a) Evaluate $$ \left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{5}\right) . $$ (b) Evaluate $$ \sqrt{1+\frac{1}{2}} \times \sqrt{1+\frac{1}{3}} \times \sqrt{1+\frac{1}{4}} \times \sqrt{1+\frac{1}{5}} \times \sqrt{1+\frac{1}{6}} \times \sqrt{1+\frac{1}{7}} $$ (c) We write the product " \(4 \times 3 \times 2 \times 1 "\) as "4!" (and we read this as "4 factorial"). Using only pencil and paper, how quickly can you work out the number of weeks in \(10 !\) seconds? \(\Delta\)
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