Difference between revisions of "Law of Cosines"
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− | + | The '''Law of Cosines''' is a theorem which relates the side-[[length]]s and [[angle]]s of a [[triangle]]. It can be derived in several different ways, the most common of which are listed in the "proofs" section below. It can be used to derive the third side given two sides and the included angle. All triangles with two sides and an include angle are [[congruent]] by the Side-Angle-Side congruence postulate. | |
− | + | ==Theorem== | |
+ | For a triangle with [[edge]]s of length <math>a</math>, <math>b</math> and <math>c</math> opposite [[angle]]s of measure <math>A</math>, <math>B</math> and <math>C</math>, respectively, the Law of Cosines states: | ||
− | < | + | <cmath>c^2 = a^2 + b^2 - 2ab\cos C</cmath> |
− | < | + | <cmath>b^2 = a^2 + c^2 - 2ac\cos B</cmath> |
− | < | + | <cmath>a^2 = b^2 + c^2 - 2bc\cos A</cmath> |
In the case that one of the angles has measure <math>90^\circ</math> (is a [[right angle]]), the corresponding statement reduces to the [[Pythagorean Theorem]]. | In the case that one of the angles has measure <math>90^\circ</math> (is a [[right angle]]), the corresponding statement reduces to the [[Pythagorean Theorem]]. | ||
+ | |||
==Proofs== | ==Proofs== | ||
+ | |||
+ | ==Proof 1== | ||
===Acute Triangle=== | ===Acute Triangle=== | ||
+ | <asy> | ||
+ | pair A,B,C,D,E; | ||
+ | C=(30,70); | ||
+ | B=(0,0); | ||
+ | A=(100,0); | ||
+ | D=(30,0); | ||
+ | size(100); | ||
+ | draw(B--A--C--B); | ||
+ | draw(C--D); | ||
+ | label("A",A,(1,0)); | ||
+ | dot(A); | ||
+ | label("B",B,(-1,-1)); | ||
+ | dot(B); | ||
+ | label("C",C,(0,1)); | ||
+ | dot(C); | ||
+ | draw(D--(30,4)--(34,4)--(34,0)--D); | ||
+ | label("f",(30,35),(1,0)); | ||
+ | label("d",(15,0),(0,-1)); | ||
+ | label("e",(50,0),(0,-1.5)); | ||
+ | </asy> | ||
− | + | Let <math>a</math>, <math>b</math>, and <math>c</math> be the side lengths, <math>C</math> is the angle measure opposite side <math>c</math>, <math>f</math> is the distance from angle <math>C</math> to side <math>c</math>, and <math>d</math> and <math>e</math> are the lengths that <math>c</math> is split into by <math>f</math>. | |
− | |||
− | + | We use the Pythagorean theorem: | |
− | |||
− | |||
− | We use the | ||
<cmath>a^2+b^2-2f^2=d^2+e^2</cmath> | <cmath>a^2+b^2-2f^2=d^2+e^2</cmath> | ||
Line 28: | Line 48: | ||
We use the addition rule for cosines and get: | We use the addition rule for cosines and get: | ||
− | <cmath>\cos{C}=\dfrac{f}{a} | + | <cmath>\cos{C}=\dfrac{f}{a}\cdot \dfrac{f}{b}-\dfrac{d}{a}\cdot \dfrac{e}{b}=\dfrac{f^2-de}{ab}</cmath> |
− | We multiply by -2ab and get: | + | We multiply by <math>-2ab</math> and get: |
<cmath>2de-2f^2=-2ab\cos{C}</cmath> | <cmath>2de-2f^2=-2ab\cos{C}</cmath> | ||
Line 45: | Line 65: | ||
===Right Triangle=== | ===Right Triangle=== | ||
− | Since <math>C=90^{\circ}</math>, <math>\cos C=0</math>, so the expression reduces to the Pythagorean Theorem. You can find several proofs of the Pythagorean Theorem [[Pythagorean Theorem#Proofs|here]] | + | Since <math>C=90^{\circ}</math>, <math>\cos C=0</math>, so the expression reduces to the Pythagorean Theorem. You can find several proofs of the Pythagorean Theorem [[Pythagorean Theorem#Proofs|here]]. |
===Obtuse Triangle=== | ===Obtuse Triangle=== | ||
+ | The argument for an obtuse triangle is the same as the proof for an acute triangle. | ||
− | ==See | + | ==Proof 2 (Vector Dot Product)== |
+ | Consider <math>\triangle{ABC}</math>. Let <math>\vec{AB}=\vec{c}, \vec{AC}=\vec{b},\vec{BC}=\vec{a}</math>. | ||
+ | |||
+ | Because of the identity <math>|\vec{a}|^2=\vec{a}\cdot\vec{a}</math>,we can complete our proof as the following. | ||
+ | |||
+ | Draw the diagram. Note that <math>\vec{c}+\vec{a}=\vec{b}</math>. Then <math>\vec{b}-\vec{c}=\vec{a}</math> and <math>\vec{a}\cdot \vec{a}=a^2</math>. <math>(\vec{b}-\vec{c})^2=b^2+c^2-2\cdot b\cdot c\cdot \cos{A}=|\vec{a}|^2</math>. Now, we have finished the proof because the two quantities are equal. | ||
+ | |||
+ | Credits to China High School Math textbook <math>\emph{Mathematics Vol 5B Textbook}</math> by People's Education Press (this textbook is currently discontinued but it has been used for hinting the proof. The proof is done by myself. But the letting and the process of guiding students to verify the identity <math>|\vec{a}|^2=\vec{a}\cdot\vec{a}</math> is written in the textbook. | ||
+ | |||
+ | ~hastapasta | ||
+ | |||
+ | ==Problems== | ||
+ | ===Introductory=== | ||
+ | 1. If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle? | ||
+ | |||
+ | <center><math> | ||
+ | \mathrm{(A) \ } 2 | ||
+ | \qquad \mathrm{(B) \ } 8/\sqrt{15} | ||
+ | \qquad \mathrm{(C) \ } 5/2 | ||
+ | \qquad \mathrm{(D) \ } \sqrt{6} | ||
+ | \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2 | ||
+ | </math></center> | ||
+ | |||
+ | ([[University of South Carolina High School Math Contest/1993 Exam/Problem 29|Source]]) | ||
+ | |||
+ | 2. In the quadrilateral <math>ABCD</math>, <math>\angle{ADC}=90^\circ</math>, <math>AB=2</math>, <math>BD=5</math>. | ||
+ | |||
+ | (1) Find <math>\cos{\angle{ADB}}</math>. | ||
+ | |||
+ | (2) If <math>DC=2\sqrt{2}</math>, find <math>BC</math>. | ||
+ | |||
+ | (2018 China Gaokao Syllabus I #17) | ||
+ | |||
+ | Solution link: https://artofproblemsolving.com/community/c4h2812553_trig__triangle_laws_how_come | ||
+ | P.S.: Since the solution is on a forum, please read all the way to thread #3 for the solutions! | ||
+ | |||
+ | ===Intermediate=== | ||
+ | |||
+ | A tripod has three legs each of length <math>5</math> feet. When the tripod is set up, the [[angle]] between any pair of legs is equal to the angle between any other pair, and the top of the tripod is <math>4</math> feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let <math> h </math> be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then <math> h </math> can be written in the form <math> \frac m{\sqrt{n}}, </math> where <math> m </math> and <math> n </math> are positive integers and <math> n </math> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the greatest integer that is less than or equal to <math> x. </math>) | ||
+ | |||
+ | ([[2006 AIME I Problems/Problem 14|Source]]) | ||
+ | ===Olympiad=== | ||
+ | A tetrahedron <math>ABCD </math> is inscribed in the sphere <math>S </math>. Find the locus of points <math>P </math>, situated in <math>S </math>, such that | ||
+ | |||
+ | <center> | ||
+ | <math> \frac{AP}{PA_{1}} + \frac{BP}{PB_{1}} + \frac{CP}{PC_{1}} + \frac{DP}{PD_{1}} = 4, </math> | ||
+ | </center> | ||
+ | |||
+ | where <math>A_{1}, B_{1}, C_{1}, D_{1} </math> are the other intersection points of <math>AP, BP, CP, DP </math> with <math>S </math>. | ||
+ | |||
+ | ([[1973 IMO Shortlist Problems/Bulgaria 1|Source]]) | ||
+ | |||
+ | ==See Also== | ||
* [[Law of Sines]] | * [[Law of Sines]] | ||
+ | * [[Law of Tangents]] | ||
* [[Trigonometry]] | * [[Trigonometry]] | ||
+ | |||
+ | [[Category:Theorems]] | ||
+ | [[Category:Trigonometry]] |
Latest revision as of 23:02, 19 May 2024
The Law of Cosines is a theorem which relates the side-lengths and angles of a triangle. It can be derived in several different ways, the most common of which are listed in the "proofs" section below. It can be used to derive the third side given two sides and the included angle. All triangles with two sides and an include angle are congruent by the Side-Angle-Side congruence postulate.
Contents
[hide]Theorem
For a triangle with edges of length , and opposite angles of measure , and , respectively, the Law of Cosines states:
In the case that one of the angles has measure (is a right angle), the corresponding statement reduces to the Pythagorean Theorem.
Proofs
Proof 1
Acute Triangle
Let , , and be the side lengths, is the angle measure opposite side , is the distance from angle to side , and and are the lengths that is split into by .
We use the Pythagorean theorem:
We are trying to get on the LHS, because then the RHS would be .
We use the addition rule for cosines and get:
We multiply by and get:
Now remember our equation?
We replace the by and get:
We can use the same argument on the other sides.
Right Triangle
Since , , so the expression reduces to the Pythagorean Theorem. You can find several proofs of the Pythagorean Theorem here.
Obtuse Triangle
The argument for an obtuse triangle is the same as the proof for an acute triangle.
Proof 2 (Vector Dot Product)
Consider . Let .
Because of the identity ,we can complete our proof as the following.
Draw the diagram. Note that . Then and . . Now, we have finished the proof because the two quantities are equal.
Credits to China High School Math textbook by People's Education Press (this textbook is currently discontinued but it has been used for hinting the proof. The proof is done by myself. But the letting and the process of guiding students to verify the identity is written in the textbook.
~hastapasta
Problems
Introductory
1. If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
(Source)
2. In the quadrilateral , , , .
(1) Find .
(2) If , find .
(2018 China Gaokao Syllabus I #17)
Solution link: https://artofproblemsolving.com/community/c4h2812553_trig__triangle_laws_how_come P.S.: Since the solution is on a forum, please read all the way to thread #3 for the solutions!
Intermediate
A tripod has three legs each of length feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then can be written in the form where and are positive integers and is not divisible by the square of any prime. Find (The notation denotes the greatest integer that is less than or equal to )
(Source)
Olympiad
A tetrahedron is inscribed in the sphere . Find the locus of points , situated in , such that
where are the other intersection points of with .
(Source)