MAT 1275: Introduction to Mathematical Analysis

Similar documents
MAT 1275: Introduction to Mathematical Analysis

7. SOLVING OBLIQUE TRIANGLES: THE LAW OF SINES

Section 2.1 Special Right Triangles

STRAND F: GEOMETRY F1 Angles and Symmetry

OVERVIEW Using Similarity and Proving Triangle Theorems G.SRT.4

The Laws of Sines and Cosines

Pythagorean Theorem and Trigonometry

ALGEBRA 2/TRIGONMETRY TOPIC REVIEW QUARTER 3 LOGS

Precalculus Notes: Unit 6 Law of Sines & Cosines, Vectors, & Complex Numbers. A can be rewritten as

PRE-BOARD MATHEMATICS-1st (Held on 26 December 2017)

11.2. Infinite Series

SMARANDACHE GROUPOIDS

Section 1.3 Triangles

Trigonometric Ratios Unit 5 Tentative TEST date

Something found at a salad bar

3.1 Review of Sine, Cosine and Tangent for Right Angles

Trigonometry and Constructive Geometry

Trigonometry. cosθ. sinθ tanθ. Mathletics Instant Workbooks. Copyright

Differentiation Applications 1: Related Rates

Similar Right Triangles

Math Lesson 4-5 The Law of Cosines

AP Physics Kinematic Wrap Up

Asynchronous Sequen<al Circuits

HYPERBOLA. AIEEE Syllabus. Total No. of questions in Ellipse are: Solved examples Level # Level # Level # 3..

CS 477/677 Analysis of Algorithms Fall 2007 Dr. George Bebis Course Project Due Date: 11/29/2007

*GMT62* *20GMT6201* Mathematics. Unit T6 Paper 2 (With calculator) Higher Tier [GMT62] MONDAY 11 JUNE 3.00 pm 4.15 pm. 1 hour 15 minutes.

MODULE 1. e x + c. [You can t separate a demominator, but you can divide a single denominator into each numerator term] a + b a(a + b)+1 = a + b

Exam Review Trigonometry

MATHEMATICS PAPER & SOLUTION

Use of Trigonometric Functions

Pythagoras Theorem. The area of the square on the hypotenuse is equal to the sum of the squares on the other two sides

1 PreCalculus AP Unit G Rotational Trig (MCR) Name:

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3

LESSON 11: TRIANGLE FORMULAE

10.3 The Quadratic Formula

Page 1

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Chapter 5 Worked Solutions to the Problems

Date Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( )

Solutions to Problems. Then, using the formula for the speed in a parabolic orbit (equation ), we have

4.3 The Sine Law and the Cosine Law

This section is primarily focused on tools to aid us in finding roots/zeros/ -intercepts of polynomials. Essentially, our focus turns to solving.

Medium Scale Integrated (MSI) devices [Sections 2.9 and 2.10]

QUADRATIC EQUATION EXERCISE - 01 CHECK YOUR GRASP

As we have already discussed, all the objects have the same absolute value of

1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Trigonometry Revision Sheet Q5 of Paper 2

Mathematics Number: Logarithms

Flipping Physics Lecture Notes: Simple Harmonic Motion Introduction via a Horizontal Mass-Spring System

Calculus Placement Review. x x. =. Find each of the following. 9 = 4 ( )

u( t) + K 2 ( ) = 1 t > 0 Analyzing Damped Oscillations Problem (Meador, example 2-18, pp 44-48): Determine the equation of the following graph.

Improper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:

Comparing the Pre-image and Image of a Dilation

6.2 The Pythagorean Theorems

Maintaining Mathematical Proficiency

12.4 Similarity in Right Triangles

THE PYTHAGOREAN THEOREM

Mathematics. Area under Curve.

Solutions to the Extra Problems for Chapter 14

MCH T 111 Handout Triangle Review Page 1 of 3

(e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer

ECE 2100 Circuit Analysis

20 Faraday s Law and Maxwell s Extension to Ampere s Law

Lecture 7: Damped and Driven Oscillations

p-adic Egyptian Fractions

On-Line Construction. of Suffix Trees. Overview. Suffix Trees. Notations. goo. Suffix tries

6.5 Improper integrals

Section 4.7 Inverse Trigonometric Functions

LEARNING : At the end of the lesson, students should be able to: OUTCOMES a) state trigonometric ratios of sin,cos, tan, cosec, sec and cot

[COLLEGE ALGEBRA EXAM I REVIEW TOPICS] ( u s e t h i s t o m a k e s u r e y o u a r e r e a d y )

Math 20C Multivariable Calculus Lecture 5 1. Lines and planes. Equations of lines (Vector, parametric, and symmetric eqs.). Equations of lines

Part a: Writing the nodal equations and solving for v o gives the magnitude and phase response: tan ( 0.25 )

12 Basic Integration in R

Lecture 5: Equilibrium and Oscillations

WYSE Academic Challenge Regional Physics 2008 SOLUTION SET

CS311 Computational Structures Regular Languages and Regular Grammars. Lecture 6

Solving Inequalities: Multiplying or Dividing by a Negative Number

Proportions: A ratio is the quotient of two numbers. For example, 2 3

GUC (Dr. Hany Hammad) 9/19/2016

Lecture 1. Functional series. Pointwise and uniform convergence.

ANSWER KEY FOR MATH 10 SAMPLE EXAMINATION. Instructions: If asked to label the axes please use real world (contextual) labels

Physics 102. Final Examination. Spring Semester ( ) P M. Fundamental constants. n = 10P

INTEGRATION. 1 Integrals of Complex Valued functions of a REAL variable

Math 426: Probability Final Exam Practice

Lecture 4. Electric Potential

Consequently, the temperature must be the same at each point in the cross section at x. Let:

5.1 Properties of Inverse Trigonometric Functions.

SSC MAINS (MATHS) MOCK TEST-1 (SOLUTION)

MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS

Math 31A Discussion Notes Week 4 October 20 and October 22, 2015

AP Calculus BC Chapter 8: Integration Techniques, L Hopital s Rule and Improper Integrals

Chapter 3 Kinematics in Two Dimensions; Vectors

Dispersion Ref Feynman Vol-I, Ch-31

PYTHAGORAS THEOREM,TRIGONOMETRY,BEARINGS AND THREE DIMENSIONAL PROBLEMS

**YOU ARE NOT ALLOWED TO TAKE SPARE COPIES OF THIS EXAM FROM THE TESTING ROOM**

Calculus AB Section I Part A A CALCULATOR MAY NOT BE USED ON THIS PART OF THE EXAMINATION

Instructions. An 8.5 x 11 Cheat Sheet may also be used as an aid for this test. MUST be original handwriting.

Activities. 4.1 Pythagoras' Theorem 4.2 Spirals 4.3 Clinometers 4.4 Radar 4.5 Posting Parcels 4.6 Interlocking Pipes 4.7 Sine Rule Notes and Solutions

PYTHAGORAS THEOREM WHAT S IN CHAPTER 1? IN THIS CHAPTER YOU WILL:

INTRODUCTION AND MATHEMATICAL CONCEPTS

Transcription:

1 MT 1275: Intrdutin t Mtemtil nlysis Dr Rzenlyum Slving Olique Tringles Lw f Sines Olique tringles tringles tt re nt neessry rigt tringles We re ging t slve tem It mens t find its si elements sides nd ngles, given sme f tem First f ll, let s see wt elements must e given Ovius, if nly ngles re given nd n sides, tis inf is nt enug t determine sides sine tringles wit te sme ngles re similr nd my ve differerent sizes S, t lest ne side must e given We nsider ll pssile ses wen ne, tw r tree sides re given s well s sme numer f ngles Mre preisely, fur ses re pssile in slving tringles: 1) One side nd tw ngles re given 2) Tw sides nd n ngle ppsite t ne f tem re given 3) Tw sides nd ngle etween tem re given 4) Tree sides re given Min tls t slve tese prlems re tw imprtnt terems: Lw f Sines nd Lw f Csines Here we nsider Lw f Sines nd te first tw prlems Lw f Sines It is ler tt in ny tringle, te igger side, te igger ppsite ngle Hwever, sides re nt prprtinl t ppsite ngles Fr exmple, in rigt tringle 30 60, if side ppsite t 30 is, ten side ppsite t 60 is 3, wi is nt 2 Lw f Sines sys tt in ny tringle sides re prprtinl t te sines f ppsite ngles In ter wrds, te rti f ny side t te sine f te ppsite ngle remins te sme fr ll tree sides f given tringle Mre frmlly, te fllwing terem is true Terem (Lw f Sines) Cnsider tringle BC: B Ten sin = sin B = sin C C

2 Prf Fr simplisity, we nsider nly ute tringle (prf fr tuse tringle is sligtly different, ut similr) Let s drw eigt t te side : B Heigt reks tringle BC int tw rigt tringles: BD nd BCD Frm tringle BD, sin = Terefre, = sin Frm tringle BCD, sin C = Terefre, = sin C Equte te ve expressins fr : sin = sin C Divide t sides f tis equtin y sin sin C nd get = sin sin C Similr rti is true fr te side nd ngle B Te prf is mpleted Lw f Sines wrks perfetly gd fr slving tringles fr te se 1) ve wen side nd tw ngles f tringle re given In tis se tringle is defined uniquely Wit n prlem we n find te tird ngle y sutrting tw given ngles frm 180, nd ten use Lw f Sines t find tw ter sides Exmple 1 Slve tringle, if = 14,! B = 40, nd! C = 75 Slutin We need t find ngle, nd sides nd 1) = 180 B C = 180 40 75 = 65 2) Using Lw f Sines, = Frm ere, using ls lultr, we get sin sin B = sin B 14 sin 40! = = 99 sin sin65! 3) gin y Lw f Sines, = Frm ere sin sin C = sinc sin = 14 sin75! =149 sin65! Finl nswer:! = 65, = 99, = 149 D C

3 Using Lw f Sines migus Cse We nsider nw te se 2) ve wen tw sides nd n ngle ppsite t ne f tem re given In tis se tringle is nt lwys defined uniquely nd we my fe sme diffiulties t slve it Tis is te migus se We will ssume tt te fllwing dt re given: sides nd, nd ngle ppsite t side Cse: ngle is tuse Tis is simple se sine nly tw ptins re pssile: tringle des nt exist r tringle is unique T understnd wy, let s drw ngle nd mrk side n its slnt side: T get tringle, we need t drw side frm te tp pint t te rizntl side f ngle Osius, if side is t srt, it will nt tu te rizntl side, nd tringle des nt exist: In rder t exist, side must e greter tn Ten tringle is defined uniquely We me up t te fllwing Prpsitin 1 Let tw sides nd, nd tuse ngle ppsite t side re given Ten 1) If, tringle des nt exist 2) If >, tringle exists nd it is unique Nte Prt 1) is ls ler y te fllwing resn: if, ten B ngle is tuse, s B ls must e tuse But tringle nnt ve tw tuse ngles Exmple 2 Slve tringle, if = 18, = 14, nd = 130 Slutin Using Lw f Sines, we ve sin = Frm ere sin B sin 14 sin130 sin B = = = 0596 18

4 Ntie, tt t tis pint we lulted sine f ngle B, ut nt tis ngle itself T restre 1 ngle B frm its sine, we n use te uttn sin n lultr similr t wt we did fr rigt tringles Tis uttn rrespnds t inverse sine We ve B = = sin (0596) 37 Nw it is esy t find ngle C: C = 180 B= 180 30 37 = 13 T find side, we n use Lw f Sines gin: = Frm ere, = sinc sin sin C sin = 18sin13! = 53 sin130! Finl nswer: B= 37, C= 13, = 53 Cse: ngle is ute s fr tuse ngle, let s drw ngle nd mrk side n its slnt side: T rete tringle, we drw side frm te tp pint Here fur ses re pssile: 1) Side is t srt t meet wit te rizntl side: Tringle des nt exist 2) Side tues rizntl side extly in ne pint: We ve rigt tringle wi is unique 3) Side intersets rizntl side in tw pints: We ve tw tringles wit sides, nd ngle : ne is ute nd te ter is tuse

5 4) Side is lng enug nd t ret tringle, side intersets rizntl side nly in ne pint: Te tringle is unique Te tp ngle my e ute r tuse Hw n we distinguis tese fur ses using te vlues f sides, nd ngle? Tke lk t tis piture In yur mind, drw side frm te tp pint Yu n see tt if <, side is t srt nd tringle des nt exist If =, we n drw nly ne rigt tringle If < <, side n e drwn n t sides (left nd rigt) f te eigt, nd we ve tw tringles Finlly, if, we n drw nly ne tringle Ntie tt = sin, s sin = We me up t te fllwing Prpsitin 2 Let tw sides nd, nd ute ngle ppsite t side re given 1) If, tringle is unique Tis tringle my e ute r tuse 2) If <, dente = sin ) If <, tringle des nt exist ) If =, tringle is unique Tis tringle is rigt ) If >, tere re tw tringles One f tem is ute, te ter is tuse Prtil wy t use Prpsitin 2 is t diretly pply Lw f Sines sin slve tis equtin fr sin B : sin B = Tree ses re pssile ere: 1) sin B > 1 Beuse sin B nnt e greter tn 1, tringle des nt exist 2) sin B = 1 We ve B = = sin (1) 90 sin = nd sin B Te tringle is unique It is rigt tringle 3) sin B <1 Let sin B = s, ten B = sin ( s) ngle B (s inverse sine f psitive vlue) is lwys psitive nd ute S, ne tringle lredy exists T understnd weter nter tringle exists, ntie tt tere is ne mre ngle wit te sme sine

6 s fr ngle B: suplementl ngle Bʹ = 180 B ngle Bʹ is tuse Suld we ept it s send slutin r rejet it? Just mpre nd nd use te ide tt te igger side, te igger ppsite ngle ) If, ten B ʹ But ngle Bʹ is tuse nd nnt e equl t r less tn ute ngle, s send tringle des nt exist ) If <, te send tringle exists ving te tuse ngle Bʹ = 180 B Nte nter wy t see weter nter tringle exists, is t lulte suplementl ngle Bʹ = 180 B in ny se (regdless n wi side is igger: r ) Ten, if B ʹ+ <180!, ept B ʹ, nd if B ʹ+ 180!, rejet it (n rm fr ngle C) Exmple 3 Let = 20 nd = 30 Determine te numer f tringles tt stisfy te given nditins If tringle exists, slve it 1) = 5 2) = 3) = 16 4) = 25 sin Slutin Using Lw f Sines =, we ve sin B = Frm lultr sin sin B (r just ntie tt 30 is speil ngle), sin = sin30 = 05, nd expressin fr 20 05 sin B emes sin B = = 1) If = 5, ten sin B = = 2 Beuse sine nnt e greter tn 1, tringle des nt 5 exist 2) If =, ten sin B = = 1 nd B = sin (1) = 90 Tis is rigt tringle T slve it, it remins t lulte ngle C nd side C = 90 B= 90 30 = 60 Side n e fund y Pytegren Terem (ntiett is yptenuse, nd nd re legs): 2 2 2 2 = = 20 = 300 = 3 Finl nswer: B= 90, C = 60, = 3 3) If = 16, ten sin B = = 0625 nd B = sin (0625) = 39 nter ngle Bʹ, 16 su tt sin Bʹ = sin B is n tuse ngle We ept it euse > ngle Bʹ is suplement t ngle B: Bʹ = 180 B= 180 39 = 141

7 S, we ve tw tringles Let s slve tem It remins t find ngle C nd side ) Tringle wit ngle B = 39 We ve C = 180 B= 180 30 39 = 111 By Lw f Sines, = Frm ere, sin sin C sinc 16sin111 = = = 2087 sin sin 30 ) Tringle wit ngle B = 141 (we use letter B insted f Bʹ)We ve C = 180 B= 180 30 141 = 9 By Lw f Sines, = sin sin Cʹ Frm ere, sinc 16sin9 = = = 501 sin sin 30 Finl nswer: Tere re tw tringles: B= 39, C = 111, = 2087 B= 141, C = 9, = 501 4) If = 25, ten sin B = = 04 nd B = sin (04) = 24 nter ngle Bʹ, su 25 tt sin Bʹ = sin B is suplement t B nd it is tuse ngle We rejet it euse < nd ngle Bʹ nnt e tuse s we mentined in Nte ve, we n ls lulte B ʹ =180! B =180! 24! =156! Ten B ʹ+ =156! + 30! =186! >180! (nd n rm remins fr ngle C) Terefre, gin we rejet Bʹ S, we ve nly ne tringle wit B = 24 T slve it, it remins t find ngle C nd side C = 180 B= 180 30 24 = 126 By Lw f Sines, = Frm ere, sin sin C Finl nswer: B= 24, C = 126, = 4045 sin C 25sin126 = = = 4045 sin sin 30