Precalculus Algebra and Trigonometry
Precalculus Algebra and Trigonometry MA 111
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MA1 1 1 Precalculus Algebra and Trigonometry Exam 1 Review Questions Solution July 12 2008 1 Which if any or all of the following relationships show a function a 1223J33J451 repeated Hence for each x value there is a unique corresponding y value This is a function An easy way to tell is to observe that none of the x coordinates are b 555664J721 This is not a function The x coordinate 5 corresponds to bothy 5 and y 6 C01141131144 This is a function for the same reason as in question a d x2 y2 1 Solve for y x2y2 1 92 17X2 y ix17x2 7V1 7 x2 Hence not a function Note that for any value of x there are two corresponding y values both x1 7x2 and eyx21 Solve for y yx2 1 y 17x2 This is a function 2 a Find the domain of the function fx x7539 Fi The underlying assumption for the remainder of the semester is that a function maps a real number to a real number So any number which does not map to another real number is not in the domain What can go wrong in a rational expression such as that above is that we can be given a value of x such that the denominator is 0 We can t divide by 0 so the output of the function is not a real number Solve the denominator for 0 If x 7 5 0 then x 5 So HE is not a real number Hence 5 cannot be in the domain Any other value for x is okay Hence the domain is All real numbers except 5 or X l X 5 or 0051U5OOJ nd the domain of the function fx x x271 39 Again lwill solve the denominator for 0 to describe what is not in the domain Letxzil 0 Thenx2 lorxi il So the domain is all real numbers except 1 or 71 or X l X ill or OOy llUl lyllUllyOOl The graph of x is given to the right a What are the xintercepts 0 The graph of x crosses the x axis atx ill x l and x 3 b What are the yintercepts 0 The graph crosses the y axis aty 3 c List quot t the relative minima 0 There is one at 2 73 d List quot fthe relative ma ima 0 Thereis one at 03 e On what intervals is fx increasing 0 fx is increasing on the interval fl 5 0 and 2 32 f On what intervals is fx decreasing 0 fx is decreasing on the interval 0 2 h For what values of x is fx 2 0 Atx fl 5 x 5 and x 31 the value of fx is2 i Whatis the value of f2 0 When x 2 theny 73 j Is the point 1 0 a relative maximum 0 No The function is still decreasing around that point Specifically to the left of this point the function is decreasing and to the right of this point the function is decreasing k Whatis the domain 0 The possible values of x lie in the interval fl 5 32 l Whatis the range 0 Thepossible values ofy liein the range 753 4 Let x x2x1 a Find fxh m fwm wh wm1 wmwmmm1 x22hxh2xh1 b Find the difference quotient for x Recall that fxhj x2 2hxh2 xh1 and that x x2 x 1 Then I will substitute x2 2hx h2 x 11 1 for x h and x2 x 1 for x and proceed to simplify fxh4x 7 x22hxhzxh117x2x1 h x XZ2hxhzXh1ixzixi1 h 2hxh2h h 7 h2xh1 7h 2Xh1 5 Let x x2 71 a Find the average rate of change from 1 to 5 Fi The average rate of change from A to B is given by Applying the formula for A 1 and B 5 we find BF AJ 7 5171 01 137A 571 2470 4 6 nd the equation of the secant line from 1 to 5 We are looking for a line Hence we start with y mx b and determine the values for m and b Recall the slope of the secant line is equal to the average rate of change from A to B Then we know how to find m We use the average rate of change forrnula We already did that above so m 6 Now I know y 6x b To find b I pick a point on the line plug the correspond ing x and y values into the equation for the line and solve for I will let x 1 I must choose either x 1 or x 5 because those are the two x values the secant line connects When x 1 theny H1 12 7 1 0 Then061borb76 Theny 6x 7 6 S The functions 8p and Dp model supply and demand of quantities of a particular brand of soup 8p 5p 7 20 Dp 30 7 75 p a Find the quantity demanded when the price is 2 Whenp 2 Dp D2 307 752 15 b Find the price when the supply is 15 We re asked to find p when 8p 15 I solve the equation 15 5p 7 20 for p Then 5p 35 or p 7 c Find the equilibrium point That is the price for which supply equals demand I m looking for the value of p such that 8p Dpj le 13le 510720 3077510 12510 50 p 4 d What is the quantity demanded at this point When p 4 then Dp DH 0 This particular brand of soup appears to be lacking something 7 Consider the piecewisedefined function below 0 ifxlt74 x x4 if74gxg4 x278 ifxgt4 a Graph the above function 48 04 40 hWanap Whopo c What is f72 H72 2 i iWanm nwm6 e What is f6 6HM 8 a Draw the graph of x x2 You should be able to check the answers to this question and each of the following using a graphing calculator Or try httpwwwwebgraphingcom httpWwwwebgraphingcomgraphingibasicjsp b State the equation and draw the graph of x after compressing vertically by a factor of a 15m 15x2 c State the equation and draw the graph of x after expanding vertically by a factor of 3 3le 3X2 l d State the equation and draw the graph of x after compressing horizontally by a factor of a f2x 2x12 4x2 l Nl 9 State the equation and draw the graph of x after expanding horizontally by a factor of 4 a m 17x12 97x2 l f State the equation and draw the graph of x after shifting left by 2 units fx21x212x24x4 g State the equation and draw the graph of x after shifting down by 2 units fx172X272 l h State the equation and draw the graph of x after re ecting over the x axis 0 4m 7x2 l i Is the point 24 on the graph of x after the previous transformation Whenx 2 then i x 7 2 72jz 74 x can t correspond to both 4 and 74 so no 9 The following demand equation models the number of units sold x of a product as a function of price p x 74p 200 a Write a model expressing the revenue R as a function of x Revenue is the product of price and units sold Hence R x p But we know x 74p 200 so I substitute 74p 200 for x in the revenue equa tion I get R 74p 200j p or R 74p2 200p b What price should the company charge to maximize revenue Note that the formula for R is a quadratic Hence its graph is a parabola A parabola has a vertex which is either a min point if the parabola opens up or max if the parabola opens down The a coefficient of the quadratic ax2 bx c is 74 That it is negative value means the parabola opens down so the vertex is a max The vertex is the point p R where p is the price and R is the revenue at that price 1 The formula for the vertex is 72 H7 Elr Sop 7 7 72 25 This means the maximum revenue is obtained when p 25 c What is the ma imum revenue We found the price p which generates maximum revenue Revenue at any price is given by R 74p2 200p So to find the maximum revenue plug in the price which gives the maximum rev enue 1225 442512 20025 2500 d What quantity x corresponds to the maximum revenue Recall that x 74p 200 Then when p 25 x 7425 200 100 9 How do you know there is a maximum and not a minimum point on the graph of this function Because the a coefficient is negative as explained in the solution to question b 10 Write a polynomial x which has the roots 72 2 4 and such that N 0 To find the other solutions solve x3 7x2 0 I can factor out an x2 from the LHS Then I get x2x 7 0 So x2 0 means x 0 x70meansx77 What is the domain of the rational function fx 7 7 the graph of x touches the x axis at 72 and 2 and crosses the x axis at 4 and has a degree of 9 If x has the roots 72 2 and 4 then it has the factors x2 x72 x74 respectively So start by letting x be the producto of these factors x X 21X7 2X4J Now the powers of the factors x 2 and x 7 2 must be even as x touches the x axis at their respective roots The power of the x 7 4 factor must be odd as x crosses the x axis at its respective root Also the degree of the polynomial must be 9 This means the sum of the three powers must be 9 One possibility x X ZJZKX 212X 415 Write a polynomial x which has the roots 171 and such that the graph of x touches the x axis at both 71 and l and has a degree of 6 One possibility x x7 l4x 1 What are the roots of the polynomial 2x 7 3 x3 7x2 lwant to find the values of x such that 2x 7 3x3 7x2 0 In other words the x intercepts So 2x7 3 0 and x3 7x2 0 If 2x 7 3 0 then 2x 3 and so x There is one solution The domain of a rational function is all real numbers except those values of x for which the denominator is 0 lsolvex26x80 x26x8 x4x2sox4 0andx2 0 Thenx 74 and x 72 So the denominator is all real numbers except x 74 and x 72 x l x 74X7 72 14 For the rational function above state the vertical asymptotes The locations of the vertical asymptotes are precisely the roots of the denominator which are not roots of the numerator So there s two ways to go about this solution We already know the roots of the denominator So we can find the roots of the nu merator and compare the lists We could also write x in reduced form and solve the denominator for zero Looking at the next couple of questions we ll need to know the roots of the numerator So let s go that route lsolvex25x60 x25x6 0 Thenx73and x72 x3x2sox3 0andx2 The numerator has roots 73 and 72 The denominator has roots 74 and 72 74 is the only root of the denominator which is not a root of the numerator So there is a vertical asymptote at x 7 For the rational function above state the horizontal asymptotes Recall the rule for rational functions 1 If the degree of the numerator is greater than the degree of the denominator there is no horizontal asymptote 2 If the degree of the numerator is less than the degree of the denominator the line y 0 is a horizontal asymptote 3 If the degrees are the same throw out all but the dominant term of the numerator and denominator Simplify to get a real number L Then the line y L is the horizontal asymptote The degrees of the numerator and denominator are both 0 The dominant term of the numerator is x2 The dominant term of the denominator is x2 2 1 PM x So y l is the horizontal asymptote For the rational function above state the zeroes The roots of a rational function are the zeroes of the numerator which are not zeroes of the denominator We found the roots of the numerator to be 7273 But 72 is a zero of the denomi nator So the only remaining zero is 73 Hence x 73 is the zero of x 17 Let pbc x3 x2 7 6x What is the degree of pbc M3 18 What are the roots of pbc What is the multiplicity of each root 3 pbc x3x276x xx2x76 xx72x3 The roots are 0 2 73 Each has multiplicity l 19 For each root state if the graph of pbc crosses or touches the x axis at the intercept Because each root has multiplicity l the graph of p x crosses the xaxis at each one 20 State the power function that the graph of x resembles for large values of lxl Take the dominant term The power function is x3 l 21 Sketch the graph of x x 7 22x lj3xx7 l You should be able to check the answer to this question using a graphing calculator Or try httpwwwwebgraphingcom httpWwwwebgraphingcomgraphingibasicjsp 22 I wish to enclose a rectangular yard with a fence I have 400 feet of fencing available a Express the area A of the yard as a function of w the width of the rectangle lknow A 1w 1 also know 21 2w 400 Solving for 11 have 1 200 7 w Then A w200 7 w 7w2 200w b Express the area A of the yard as a function of 1 the length of the rectangle Instead of solving 21 2w 400 for 1 solve it for w Then w 200 7 1 So I can get A 712 2001 c Find the maximum area Find the Iquot of the rectangle which give this area Given A 712 2001 is a quadratic its graph is a parabola So it has a vertex The parabola opens down because the leading coefficient is negative So the vertex is a maximum The vertex is the point 1 A where 1 50 100 Knowing 1 100 andw 20071thenw 2007100 100 The dimensions are w 1001 100 When 1 100 the area is 712 2001 71002 200100 10000 23 A cylindrical box is to the constructed such that the sum of the height and radius is 100 inches Construct a function which states the volume of the box as a function of its radius Knowing the volume of a cylinder is V 7TT2139L lhave a model which takes r radius as input and gives me the volume But what is h This is an unknown To eliminate it I need more information The sum of height and radius is 100 So 11 r 100 Solving for h I get 11 100 7 T So now I can substitute 100 7 r for h in my volume formula 7T r2100 7 T Now I have a formula for volume whch is a function of T as was requested 24 Which polynomial below is that whose graph is depicted to the right Choice A gel 3 Choice B To Choice C gel xi 3w Choice D 7 x731x Choice E 7x3x731x753 x73x75 3X53 X 75 75 3 1 The correct answer is A The graph has roots 0 3 and 5 eliminating choice C The graph crosses thex axis atx 3 eliminating choices D E The graph crosses thex axis atx 5 also eliminating choice D Choice B is eliminated because the graph touches the x axis at x 0 but the power olxis odd 25 lnthe graph above whatvalue does x approachasx approaches x c x a co 26 Which polynomial below is that whose graph is depicted to the right Choice A 7x il1x 23 Choice B 7X l1x 23 Choice C 7x r l Zx r 2 l Choice D 7x r l 3x r 23 Choice E 7x r l1x 72 The correct answer is E The graph has roots l and 2 eliminating choices A B The graph touches the x axis at x l eliminating choice D The graph crosses thex axis atx 2 eliminating choice C 27 Fill in the blank In the graph above as x gt foo x gt 7 x gt oo 28 Using transformations of the graph of the function x 1 sketch a graphof gx S 1 You should be able to check the answer to this question using a graphing calculator Or try httpwwwwebgraphing com httpWwwwebgraphing comgraphingibasic jsp 29 For gx as defined above fill in the blank As x gt 4 gx gt l x gt foo N MAl l l Precalculus Algebra and Trigonometry Exam 3 Review Questions Solutions August 5 2008 1 Give an example of two angles 61 and 62 such that 00561 sin62 Remember that if 61 and 62 are complementary their sum is 90 then 00561 sin62 Why is that So I can pick any two values that sum to 90 I ll pick 61 30 and 62 60 Consider the following trigonometric questions a Convert 60 to radians m g b Convert 771 to degrees l e 7180 c Compute From the Complementary Angle Theorem I note that sin73 cos 7 So cos 7 2050 70 1 d Compute the length of an arc which lies on the unit circle and subtends an angle of l radian Citing the definition of a radian I can read the length is l If I didn t recall this definition I can still find the value using the formula 5 T6 where r 1 because the arc is on the unit circle and G 1 Remember that G was defined in radians not degrees I obtain slgtltll Lz mm mmmemaze mum m wan hadnsas39 msmy b mm 7 mmmkw m mmmw mwwgamamw n MANN ham 5 mm by m cmmnm W m memo AIEI mozmmdAzsmmar 90735 we 5 mass mum Emhlmnsmeamgauvelugm dnesnm mu amzdmthSMd mmhmmu a mn ms 3 Wu K Duymsuyywe mm m magnum mm m m 5 was on m a fig q mc3A Lue z gmvbmmmcm i m um Ems 7 m am m y Mama 1 g m mem a I mmm Vmse sume1y nerquot 4 L mmommdsmm m ammunsmmonmg g 9 X 242 7 re Do mnme m ms was mm 51 an Lu my afamommwm mm mmew J K39 m561 7 J 0 rummammmmw wh n xsh mmduuudzhAyqu mygmm amgm Duyuunmbuwhy sawssz 7 7 mm 111257 mum uluxlzlAandklww swung g g mommumsmmm g 5me 4 Give examples of each of the following a A sinusoidal equation through the point g l From the unit circle I know when G sin6 1 So I pick sin6 as my equation b A sinusoidal equation with range 1 5 I know y sin6 has range 711 If I multiply y sin6 by 2 and shift it up 3 units I ll have the range 15 So how did I know to do that Remember that when establishing standards for sinusoidal modelling we derived A 15MAX 7 MIN and lZUVlAX MIN The min point in the range 15 is l and the max point is 5 So B A2andB3Solgety2sin63 c A trig function with range 700 71u1 00 lrecally sec6 has range 70071Ul oo d Transform your function above so its range is 700 0U2 00 All I need to do is shift my function up 1 unit If you can t1picture that in your head you can also use the formulas A 1 UVlAX 7 MIN and B 5MAX MIN again In this case you would get A l and B 1 Can you figure out why the formula for A still works 5 Determine the exact value of each of the following 13 a sinT lwill use the period properties of sin to figure this one out Recall sin6 sin6 271 sinl3T7 sin13 TTE 7 271 sin7377 sin7377 7 271 sing Elh b tan30 gtlt cot30 C sin9 cos939 Recall tan6 And so cot6 l S s Therefore m9 2059 1 sin9 tan6 gtlt cot6 gtlt Notice that this is true for G in general So it really doesn t matter that G 30 The same answer would be obtained if G 60 or even if G 351524352quot The only exception is if 6 isn t in the domain of tan or cot If you didn t catch onto that you could still have determined the values of tan30 and cot30 using the unit circle and obtained 1 cos15 39 2053450 This was one of the A questions because to get it you really had to know your trig function properties Because the period of cos is 271 or alternatively 360 1 now cos345 cos345 7 360 cos715 Also because cos is an even function cos715 cos15 Last cos15 7 fuse The alternative approach would have been to use the halfangle formula for cos to compute cosl 30 to obtain cos15 and then use that result to compute cos330 15 but the algebra would quickly get out of hand d sin W To solve this ljust look for what 6 gives sin6 There are two solutions 6 Use the given information to find the exact value of the following a Given cos6 7 and G is in Quadrant H Find sin6 Since I know cos6 and the Pythagorean Theorem I can easily obtain the absolute value of sin6 sin26 cos26 1 sinztej 7312 1 sin26 215 1 sin26 12 sin6 ig Now I need to pick either or 7 Since G is in Quadrant II sin6 should be positive So sin 6 3 b Given cos6 715 and sin6 7 find tan6 Recall tan6 39 so cos f tan6 3 73 J gtlt 2 c Given cos6 g and 0 g G g find sin26 Using the appropriate angle surn identity I know sin26 2sin6 cos6 I already know the value of cos6 and I can obtain the value of sin6 in the same manner as in part a In the exact same manner as in part a I cornpute sin6 i5 Because 6 is in Quadrant I I know sin6 must be positive so sin6 3 Therefore sin26 2sin6 cos6 2 X g 7 Given y 74 sin3x 7 2 find the amplitude period and phase shift of the sinusoidal curve Consider y Asinwx 7 ch lAl gives the amplitude so A l 7 3 3 The period T 2f 2 The phase shift is given by 8 Compute the exact value of the following a sin 1 sin e Sinc 7 g g g then sin 1 sin b tansin 1 715 Let s begin by first thinking about the value of sin 1 71 G This expression yields 6 such that sin6 715 From the unit circle I know 6 7 I don t consider the other possible value since the range of sin 1 is 7 g wl l M4 sin7 cos7 J 7 71 7 2X 1 a So tansin 1 715 tan7 wl M c cos26 given sin6 g and G is in quadrant ll A quick way to do this is to recall cos26 l 7 2 sin26 Then plug in g for sin6 9 Find all solutions to the following a cos26 7 Let u 26 cosu 7 has two primary solutions for u They are 5quot and Now recall the period of cos is 271 This means cos6 cos627 r cos647 r cos62k7r where k is an integer So we arrive at the following u52k7r u2k7r 2652k71 26767quot2k71 6 7 k71 9k b cos 7 37quot This is much too difficult to be solved without the use of a calculator so it was dropped Can you solve this with a calculator to help you 10 The following table gives the average monthly temperature in South Watsonville Show how to con struct a sinusoidal model y Asinwt 7 ch 4 B modelling the data as a function of t temperature l3l4l5l6l7l8l9l lMonth llanl2 l9l44l55l61l71l77l84l71 Temp F l 76 a Find the value for A 9 As discussed in class A can be found by A 15MAX 7 MIN This gives A 46 b Find the value for B To find B consider B 1 UVlAX 4 MIN This gives B 37 c Find the value for w Consider the period as 12 Then 12 Solve for w to get u g d Find the value for d The partially completed model is y 46 sint 7 Cb 37 This leaves I as the only unknown To obtain its value we must solve for it by substituting values for t and y Do you remember the trick to make this computation relatively easy I pick the min data value 179 to substitute This gives 79 46 sin 7 ct 37 746 46sin 7 ch 71 sin 7 ch Now 7 d is an angle Let s call it u Then 71 sinu We can easily solve for u by referring to the unit circle It is So 3 u 7 means that 3 7 7 71 Now I can solve for d 3 7 77d 87 7d d 74j So the completed model is y 46 sint 4Tquot 37 n lsmunpnm gs mm mummy 3 m madam m m M aim amth 5 a mmph ulwmmbeem A may zmsmm mum 4 mm m shmdd m ugly dAvAnnnAd by mm mm I may my mum Mm m lax1nd m Wu pm W yd me 59 m Wm v s w r m m m Eu MAximle mm in x mm m yamz y muplAvAd WM 5 y nosmguduvnghmwmm Ham 5 y omsm 0 1 WM 1mm mm mm NW 1 Wm m Wu whmwmum Mm may Md u at 533 Solublm 12 Suppose sin6 The identities to find the exact value of each of the five remaining trigonometric functions cos6tan6csc6sec6cot6 Consider a right triangle The sine of an angle is the ratio of the lengths of the opposite leg to the hypotenuse So we can say the opposite leg has length 3 and the hypotenuse has length 5 To find the length of the adjacent leg use the Pythagorean Theorem 32X2 52 9x2 25 x2 16 x 4 So the length of the adjacent leg is 4 sin6 fse cos6 tame h 3i 212 0509 sinle 3 sec6 wslw cot6 1W g 13 Suppose tan6 The identities to find the exact value of each of the five remaining trigonometric functions sin6 cos6 csc6 sec6 cot6 Again consider a right triangle Tangent is the ratio of the opposite leg to the adjacent leg We consider the opposite leg to be 3 and adjacent leg to be 5 To get the hypotenuse use the Pythagorean Theorem 32 52 X2 9 25 x2 34 x2 m x2 7 opposite 7 i 51116 7 hypotenuse 7 34 7 adjacent 7 5 cosle 7 hypotenuse 7 W 7 adjacent 7 sin9 7 7 3 tame 7 hypotenuse 7 6059 7 7 5 csc6 m secle Tisha cot6 mule g g Without a calculator determine the value of sec283 7 tan283 Recall thatl tan26 sec26 So sec283 7 tan283 l tan283 7 tan283 1 tan283 7 tan283 1 15 Without a calculator determine the value of cscz 3 7 cotz 3 j gtlt cosl 3 gtlt secl 3 Recall that l cot26 csc26 Also recall that sec6 519 So csc2l3 7 cot2l3 gtlt cosl3 gtlt secl3 1 cot213 409030 gtlt cosl3 gtlt secl3 l X cosl3 gtlt secl3 1gtlt cos13 X W 1 X1 1 16 Without a calculator determine the value of SSEL 3 J in 57 39 Because 33 57 90 and cos and sin are cofunctions then cos33 sin57 cos33 7 sin57 71 sin57 rave 39 So I write 17 Without a calculator determine the value of cot10 7 cosq 00 Sam 00 ltgt 3383 7 comm so cot10 7 333 o 18 Given cos30 l determine the value of a cos60 In class we derived b 51112800 We know sin230 cosz30 1 cos30 so cosz30 jz i 4 Then sin230 1 sin230 sin30 15 C sec30 3 594300 cosl gt0 A d csc30 By Complementary Angle Theorem sin30 cos60 Then csc30 m 2 19 A ship offshore from a vertical diff known to be 900 feet inheight takes a sighting of the top of the diff The angle of elevationis found to be 30 How far offshore is the ship 0 We want to find X the distance from the diff Iknow tan30 gg so 577 xe Solve for X to get x m 155979 20 To measure the height of the Watson Towers two sightings are taken at a distance of 120 feet apart If the first angle of elevation is 66 and the second is 60 what is the height of the tower X 120 W The trick here is to look for right triangles There are two Both share the leg formed by the tower vertical red line But the base of one is the horizontal red line with a length of x The base of the other is the blue line with a length of 120 x The key is the tangent function tan6 I can get two relations tan66 2246 h x tan60 1732 mg l have two equations and two variables This gives me a system of two equations 2246 eqn 1 1732 Tzloler eqn2 Solve equation 1 for x Now plug this solution into equation 2 7 11 1732 712mm 1732120 2 Q46 11 20784 77lh 11 20784 22911 11 9076 21 A point in the terminal side of an angle 6 is 2 3 Assume the vertex of the angle rests on the origin Find each of the following a sin6 The origin the point 2 0 and the point 2 3 form a right triangle The line along the x axis has a length of 2 Call this line a The line from 2 0 to 2 3 has a length of 3 Call this line b Thencz22324913socm sin6 g b cos6 cos6 C tan6 tan6 d csc6 csc6 m ms e sec6 a sec6 051 m2 1 cot6 mode 23 g G in degrees and radians tan6 b 3 so 6 tan 1 g m 5631quot 3 22 The hypotenuse of a right triangle is 10 inches One leg is 4 inches Find the degree measure of each a le 0 We just need one o the angles Call it 0c Because the sum of the angles of a triangle is 180 and the triangle is a right triangle the other angle will be 90 7 0c Lets find the angle 0c that is formed by the hypotenuse and the known leg The known leg is adjacent to the angle so well use the cosine relationship ada ent C C051 m 5 cosoc 7 Then 0c cos 1 1 m 6642 Then the other angle is 90 7 664Z 2358 Milton is sitting on a wall 4 feet above the ground He is watching a squirrel in a tree 15 feet away If the angle of elevation is 55 how tall is the tree N 9 15 0 Look at the diagram We can use trigonornetry to find h Iknowtan55 so adacent 1 428 Solve for hto get 11 15 X 1428 2142 But the tree is an additional 4 feet tall So the height is 2542 24 Solve the right triangle shown for each of the given parameters ET X 4b5 l To get c use the Pythagorean Theorem 4252 02 1625 02 41 c2 m c opposite To get 5 note that sinl W W Then 5 sin l m 896055385radians 51342 To get a note that the sum of the angles of a triangle is 180 So a must be 180790751340 3866 15 0 5 l To get 5 note that the sum of the angles of a triangle is 180 So 5 must be 1807907150 75 7 adjacent Now note that coslfl 7 hypotenuse So cos75 Then c W350 m 1932 7 opposite Now note that sinl 7 ihypotenuse So sin75 T b 3239 Then 19 1932 sin75 m 1866 C oc40 B 500 There are multiple solutions to this problem Here is how to derive one First note that sin50o 7 hopp l so ypotenuse 766 b Q There are many choices for a and b Hence why there are multiple solutions to the problem The easiest is to letb 766 and a 1 Use the Pythagorean Theorem to get C 7662 12 CZ 586756 1 CZ 1586756 CZ 1260mc 25 A point in the terminal side of an angle 6 is 3 4 Assume the vertex of the angle rests on the origin Find each of the following a cos6 g l ET tan6 m g l csc6 3 l P d sec6 g l cot6 g l 5 N O Milton is sitting on a chair It rises 2 feet above the ground He is watching a squirrel in a tree that is 19 feet away If the angle of elevation is 42 how tall is the tree 19tan42 21911 l 27 A right triangle has legs of length 2 3 What is the length of the hypotenuse What are the measures of the angles Hypotenuse m Angles 90 tan 1 g m E63l tan 1 g m 33690 28 A right triangle has one angle of measure 19 What are the measures of the other angles and legs Angles 90 l9o 71 There are multiple solutions for the legs Assuming the hypotenuse has length 1 then Legl sinl9 m 326 Leg 2 sin7l m 946 Notice that 3262 9462 m 1 We should get that because of the Pythagorean T eorem gt0 If c056 I and 6 lies in quadrant IV find 51n6 and tan6 Iknow that 51n26 c0526 1 So this gives me 51n26 51n26 210r 16139 Now I solve for 51n6 51n26 117 1 51n26 117 1 51n26 51n6 i S015 51116 M 0174 Since 6 15 in Quadrant IV then the sine of 6 15 negative Recall the starred chart from class Then 51n6 7 tan6 372
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