unit 3 - right triangle trigonometry -...

Unit 3 - Right Triangle Trigonometry - Classwork

'-- We have spent time learning the definitions of trig functions and finding the trig functions of both quadrant andspecial angles. But what about other angles? To understand how to do this, and more 'importantly, why we doit, we introduce a concept calledthe unit circle. A unit circle is a circle whose radius is one.

To the left is a unit circle. The angle e is drawn in the first quadrant but could bedrawn anywhere. Suppose e = 40°. If we were to find sin 40°, we know that it

would be defined as Y = y. So when we take sin 40° , we are finding the height1

of the triangle in a unit circle. The same argument holds when we take cos40° ...we are actually finding the x variable in a unit circle. When we take tan 40° , weare finding the ratio of y to x in a unit circle.

On your calculator, be sure you are in Degree Mode and set your decimal accuracyto FLOAT. Use your calculator to find sin40° and cos40°. Remember what it isyou are finding: the y and x variables in the triangle above. And sincex = cos40° and y = sin 40° , let us show that the Pythagorean theorem holds in thistriangle based on the unit circle.

iSl n(41::1._4'''''''8-:0·- ''''9~• '=' ..::.( ('='''' ("-os(413.7660444431(sin(40»<:+«cos(40)21

Taking trig functions on the calculator is straightforward: type in the trig function (you will get a leftparentheses) and the angle. You do not need to complete the parentheses. Press ENTER and out it comes.Although we can get extreme accuracy, we will find that four decimal places is usually enough. So set yourcalculator to 4 decimal places. Rememberthat angles are assumed to be in radians unless in degree format.

Example 1) Find the following:

a)sin29°= O,~SL.{S b)cos131°=tO. &s""(gC} c)tan(7;)= (-0S.I..II'i2.)

TAJ(' .4~")If angles are input with more accuracy, it is assumed that they are in decimal degrees. Note that parentheses canbe used to make the problems clearer in intent.


a) tan12.8° = ,0. ~'2.-, (

If trig functions of angles that are in degree-minute-second form, use the Angle menu toinput them. Remember that seconds are input with ALPHA +. cos38°40'29" would beinput to the calculator thusly:

tOS(3804"'29" I.78137


a) sin82°12' = O. ~ ~07

S~(82 ~ I~)3. Right Angle Trigonometry

b) cos126°42'53" = (; 0, ~q 7~) c) tan(-8°57'16")= (-O,/S' 75")

~~ '2.(P ~ t;} ~ ~C:oJ ~ l- ( ~ -+- :: + 13"'-c 0)- 1 - www,'mastermathmentoLcom - Stu Schwartz

Note that there are no keys for the csc, see, or cot functions on your calculator. Tofind them we have to use the fact that sin and csc functions are reciprocals of eachother, as are the cos and see functions, and the tan and cot functions. There are threeways to find, for example csc S?". Take sin S?" and then take its reciprocal or simplyfmding 11sin 3T', The screen on the right shows these two methods. You could alsofind sin 37° and then press the reciprocal key X-I.

isln(3(l/Ans1/sin(37

.60181.66161. E.61E.•


a) csc 81° = I, () I 2.ti

d) sec338.292° = I, 0 '1~3

b) sec122ci=f- /. B87D) c) cot34.2°= \.Y'1IL../

e) cot14°29'36" = 3.8b85" f) csc149°50"= I,qt.J'l3

Many times, we want to reverse the process. We know the sine of an angle and we wish to find the angle itself.To accomplish this, we use inverse trig functions or arc trig functions. These are found on your calculatorabove the sin, cos, and tangent keys. We use the blue (2nd

) key to input them ..

For instance, let us find the first quadrant angle whose sine is .7523. Note the screen onthe right. Our answer would be 48.79° (expressed in decimal degrees). Ifwe wantedour answer in degree - minute - second format, note how we would accomplish that byusing the Angle menu:

.'~8.7900

Example 5) Find the following (decimal degrees):o

a) sin' .9099 = fo S. Y <it c;o 0

b) arccos0.4231 =fs,Y. 9'95 c) tan" 1.8089= 'I. D0 ~2

Example 6) Find the following (Degrees - minutes - seconds)o I ,/ e I II 0, ~I

a) arctan4.002 = 75 5'8 /3.7 . b) sin-'.0809 = 'I 38 2 ~ 07 c) cos-1.4998= ~ 0 0 L/Z b

Finally, if we wish to find an arccsc, arcsec, or arctan function, again, there is no onekeystroke that will give it to you. To find csc' 2.3552, for instance, we must first take thereciprocal of2.3552, and then take the arcsin of that. On the right is the way you wouldaccomplish this (with the optional changing into degrees-minutes-seconds):

.424625.1249

Example 7) Find the following (decimal degrees):_I I 0 •• tJ

a) sec " 1.76 = COJ 1.1",=5'f).?/lloL/ b) arc cot 3.4221 = 1~,1.gq3 c) csc'" 1.1102= roLf.25'53, I I

I<\n.<--r~,j ~ 'iZ:Z.I ~ _

Example 8) Find the following (Degrees _ minutes _ seconds) I. II 0Z

. 0 , II b I " 0, /1

a) arccsc3.8621 = J§ {) 22.9 b) coC'0.7S01 = 53 735.1 c) arcsecS.8621= 80 /0 ,-/0.9

About errors:

Your calculator can take trig functions and arc trig functions to extreme accuracy. However, if you input theproblem into the calculator incorrectly, one of two things will happen. One - an error. Assuming you typed itin the correct syntax, the calculator is telling you that it cannot perform the operation. This is actually good for

3. Right Angle Trigonometry - 2 - www.mastermathmentor.com - Stu Schwartz

you. For instance, if you take cos' 1.4231, the calculator gives you a domain enor because we know that ther>. cosine of any angle cannot be greater than one. Hopefully, you would be smart enough to realize what is

happening. But if you input the problem incorrectly into the calculator and the 'calculator can perform theoperation, you will get an answer and you will believe it. For instance, suppose you wanted to find coel 3.5.The correct way to accomplish this is on the screen below on the left. But many students know that areciprocal is needed and do the steps on the right. They write down their answer of 0.01350 and they neverknow theyare wrong. That is why it is vital that you learn these steps now and learn them well.

l\..an-1<..••••• 574.0546l/An:=. •~3135

right way of calculating coC13.5 wrong way of calculating coC13.5

Prove to yourself that sin and sin' are indeed inverses by taking the sine of an angleand immediately taking the inverse sin of that answer. Or reverse the process. In bothC<ilS~A;;,~~.w4·eJ1d up with.the value you started. Note that you must use the Anskey on your calculator to achieve this effect. Type in the value and you will not getthe number you started with because of round-off error. This is a good way todetermine whether you truly understand how to take trig functions and inverse trigfunctions using the calculator.

Sln(:::::'sin-1(Ans 85.0000sityl(.9962 85.0035

Solving Right Triangles:

If w¥ know two pieces of information of a right triangle, we can solvethat triangle - that is find all the missing information from that triangle.For this section, we will make some generalizations: that the triangle willbe labeled .-1ABC where the side opposite angle A is lab~ded a, the sideopposite B is labeled b and the right angle is angle C and the hypotenuseis labeled c.

lJ

~

/ I"/ ./ !

A L-.._ .. .._' •.•.•__..~ - __ _.IJ c.

We can solve triangles if we are given a) an angle and the hypotenuse, b) the angle and a leg, c) a leg and thehypotenuse; and d) two legs, OUf tools will be our trig functions and the Pythagorean Theorem: a2 + b' = c',

You should always start by drawing a picture of the given information. While it doesn't have to be to scale, itshould be somewhat close. Work should be shown and answers placed in appropriate places. We will assumethat accuracy in sides is two decimal places and angles are to be in decimal degrees unless otherwise stated (if aangle is given in degree-minute-second format, assume t,hat the other angle should also be in that format).

To best understand the process, expect problems to be given in this format: We will draw the picture and do thework on the right side. While the calculator does the number crunching, you must show how you arecalculating the answers. ===-

!5o!-lCI)H7D/l

I

A=

~:-90-0---b= ~~<,

c.. 0-- 13 C J.r PIIfA)' 16 rf=' l3 -;> ,Q

~tt~y ~/-'I-¢":"-Gt--c=~A

GL;: Co .svi1. A

a= _

c=

3, Right Angle Trigonometry - 3 -

Jr:: c, .•..<.M.. B

LG,'n"l2.!', Wl'1,G~ WI~INC;

www,mastermathmentoLcom - Stu Schwartz

Example 9) Angle and hypotenuse

A =21° a= r. ()IIIB= foqO b= \3_ 1:11 b\

C =900 c= 14

~QN~

~C Jr A

0~ qO-'l.}';. ~'I

~ :0 stht-A ..& :0 ~,:\,,-! l"i

Q. -e, It.f scM ~0 h- •• 14 ~C2IJ

Example 10) Angle and leg

A = 77.20o

B= 12. eC =90°

a = 29.5b= (P,loZ'Z.. ..c= 30iZ?18

~

1~c, 0.:= '2.(t .; 6

G -: 9Q - 77. '2..-:;-IZ. ~

Example 11) Angle and leg

A = 38°12'44" ®o I II

B= Sl W7 IlD

C=90°

BQ"I!)~''5S'~

C. ,J,- 'f't

Note that we can find B:B q s Gj ~o ~ = ~ ~3<01'2.41/ Jv Q

-51 41 ,~ Ar=- Qk~

a = 102.35

b= 136.00'7c= /fe:,>.tfc,/

Example 12) Leg and hypotenuse

A= 37,9719'"B= ~2. 01-CI

C =90°

a = 8

b= 10. Z'f7c= 13I l.q

r..i./~ I\.l" -rtt,P,U! (3u7

G... c k- }r= {Io:?" J

~,:~Ac,-IQ

A-=. 1;'.'__ --"'> ®., _'I.. c..

Example 13) Leg and leg

A = :77. 9 9 'i(,e""B= 32. DD';7i.f

C =900

Ak--:;I$~~

a= 2feet='24cr.. L~B0",='2~

b = 15 inches1/

c= 2~.301"

Angles in degree-minutes-seconds

3. Right Angle Trigonometry - 4 - www.mastermathmentoLcom - Stu Schwartz

Multi-Step Problems:

Example 14) Consider the picture below.l want to find the length of segment AB. SupposeLA = 25°, LB = 40° and BC = 12. Do the necessary work on the right to find AB.o ..

?' /I) LIl!>CL c.) h4. 't eo

/./ - 90,4' .<;0 U!.'" ~ e IJ mes-a»/' / \, FI"\ C (..\) "'.., CI~' e " I'i1 (;"'r be..

//., ® 2.. ~: ~ 'l> 0 Pt _ ,,-C® ,4" .1140\) 40°· ~~ ~ 21. S"9 ® c.- ~ 1~

B \'2.. C ® \"1-3. f\-r3 z: A t: - (3c -:: 9,:; 9 J'f

Dc... L ~'12 =~L.fo.

'D c. ==- 12. ~ 1.(0 ®- I\). e b9/

CDRC.

co -= ~.18 0

AC-7b->J~€ Jo flfJ L E M eo r: r;o-« .r-tJ Il r1r;F1l.

;l)()u5) 1r 13 ~o:Sr' At. ~(?~!L/I t:"4({)M

Real-Life Applications

Guidelines for solving a triangle problem:

a) Draw a sketch of the problem situation. Don't be afraid to make it large.b) Look for right triangles and sketch them in.c) Mark the known sides and angles and unknown sides and angles using variables.d) Express the desired sides or angles in terms of trig functions with known quantities using the

variables in the sketch.e) Solve the trig equation you generated and express the answer using correct units.

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Surveying Problems: Problems involving finding quantities that would be too difficult to measure using rulersand other instruments.

17) I am standing at point A on one side of a river and wish to measure the distance across a river to a houseH on the other side. I walk 50Jeet along the riverbank to point B and measure the angle ABH to be18°23~ Find the distance across the river (assume a right triangle).

I1ll

k::. ~/,fcz/-;;.1-.,.-k z: s» 0-- (107- ~~).:::. Ire" ct b~ {Jt,

18) A guy-wire must be attached to a 50 foot pole. The angle that the guy wire must make with the groundhas to be 75.50

• Find the length of the wire required to do the job.

Angle of Elevation and Depression. As a person at point A looks upat point B, an angle of elevation with the ground if created. When Blooks down at point A, an angle of depression is created (with thehorizontal). These angles are congruent. Why?

Ilt..--r lloj, L.s Artr: ~ . Angles of elevation and depression arealways created with the horizontal.

19) A device for measuring cloud height at night consists of a vertical beam of light that makes a spot on theclouds. That spot is viewed from a point 140 feet along the ground and the angle of elevation is 64040'.Find the height of the cloud.

A -e. /II 0 'ti..,~f"'flQ r)h :=. 295.727/r:

20) While standing on a cliff 163 feet above the ocean, I see a sailboat in the distance at an angle ofdepression of 21.20 What is the horizontal distance to the sailboat?

3. Right Angle Trigonometry - 6 - www.mastermathmentor.com - Stu Schwartz

21) A wire holding up a 40 foot telephone pole is 38 feet long. The wire attaches to the telephone pole 10.5feet below the t08' What is the angle of elevation of the wire?

] 10,5' C3G -= Lfo-Io.~ .•• 1.J;,SIrz I C' - f\ (3.e. 2Cj,S

f Q 4 t\ ~ (.M.. l"t == - ::. -'J~ (..)<3 38

vi = pUy..-1 ~ =- GO. q '2.4 'I- 01\ 38 ..

c22) From a window in building A, I observe the top of building B across the 50 foot wide street at an angle

of elevation of 74°25'. I observe the base of building B at an angle of depression of 52°18' . Find the CALCheight of building B.

A

I

C3D= 5"0 7Lf2.~/,; 14.</cO® 52°'8'= 5'2.3@

~ ~ ~ls-2ol'a')r3D @Be.. =: 01> (~ )2c\~I)

=- S-o ~@ ·:daY,(D? @>o

~ :7 -bvx 6L/1.~I)GO @AB-= {3D (k-,L{ °2~/)

'= SO (t&t@) = II q. 2-S @

5"'b'Bearing and Course: When.ships or planes navigate, they need to have a simple way of explain whatdirection they are traveling. One way is called bearing. The bearing will be in the form (N or S angle E orW). A bearing is always drawn from the nearest north or south line. A heading (or course) is always drawnf h h line i I k . di . F 11' hi di . h beari drom t e nort me m a c oc Wise irection. 0 owing are s IP irecnons, t e eanng an course.

",,;:t

6Y' 72'G-IH'!/'

'Cf•.•~

",2(}:" ~.""

'it;. I".! k"'=~

\\~ .~•• " 1,tJi'TOLJAQi)S '\Q ~ Ail.OS

V vBearing: N600E Bearing: S700E Bearing: S35°W Bearing: N72°18'WCourse: 60° Course: 1lO° Course: 215° Course: 287°42'

Tip: in word problems, whenever you see or are asked for a bearing or heading, always look for_theword"from" and draw an x-y axis at that point.

Tip: if the bearing from A to B is N65°E, the bearing from B to A will be the same angle but the oppositedirection: S65°W.

23)Ajeep leaves its present location and travels along bearing N62°W for 29 miles. How far north andwest of its original position is it? I 0 _ 0 ®-

tJ, No Yl,"., -:: S. ~ '28 tVeJl.:r4=- '2 ~ Il!M 2.~-:;\3. ~ (Y)t(,.,Gt\lD JJvlt#ST,-~q "q"".. .

~ ~ L~ 0 @\N ~ q~ ~' W~~T 6.J?2Bo w~sr~ 1Ji6r:>2H "" 25. ro ~fl <.-i 'lq f\1\~

I S;-~It'i ChECK 411Pr(tlJyOlleflrJ V@ ~Z1 ~ 29 ./3. Right Angle Trigonometry - 7 - www.mastennathmentor.com - Stu Schwartz

24) An airplane leaves an airport and travels due east for 255 miles. It then heads due south for 330 miles.From its current position, along what heading should it travel to reach the airport and how far is it? ~Bur --fl ~ a.r: ~1..AAJf C)o-r 'l4EftG !I!.11.~4LJy viI( j)6Ef 1/ /J1l4r-r/?/t ?/I f.J- l 1-1-11( o1(1) 11 - J IIs r !}) s or/; t? Pit() d Ls:/J1 /

'Z '2> _ -L.8330 - \..a/lt

1-_ -I '2.is 0 a-.6- ;. lam ~3(J -== 3'-1. U ~l- 0 / . 11\

-=St} 'f 17 I '2 ,7fo fi

e330

2'Z.S

25) Two small boats leave an island at the same time. Boat A travels due North for 21 miles and Boat Btravels due west for 18 miles. How far apart are the boats and what is the bearing of boat A from boatB? How about the bearing of boat B from boat A?

N BoAT t3 ~rt..oM ~Dtrr A

V·~ ""-i1( ~~~+:~~~"-1'''''' ,~ )" .'Ie '10·"') lU

"'-----4

26) A plane leaves an airport and travels 2 hours along heading 120° at 175 mph. It then turns onto heading30° and travels 2.5 hours at 200 mph. How far from the airport is it and from its last position, what isthe heading to the airport? .J. "",,~ () •.•. _ .

~. 1 = h -c ~ :; liS ho"'- "2. ~ J So 1\1\,,,(.() \J tt ~E CA -''-' I+f3 (-I D I,.J~ -rc

A-IR.. p~ f't r d,. = '2.00 M.~ l'2.. ~) = 5"00 (Yh.'"

v.fl.

o ./'1'2.0 ..-

(3l.. r- 13- C {5 ::: '2 5'"o\J3 - 1'1 5"

A D = '2 S"~ ®Ac + FIE- z: l?sfi -t- z.,.,-o

DI':;T, /0 .41r'2PoVLT' -=. 55:3 ®== ~ @)~)'L+~r:;)i:''=- ~IO,32;rn: r>.

h 0 .i-. -I~AI)®) _ 2.'-1;'" ,. eilDIAl1 =- (~() + (;Q./)1, 'eO ~ ® - .

- 8 - www.mastennathmentor.com - Stu Schwartz

\ Iv\ P EI 1l.1'" c. ~ ,....

i'k,~ I~ A

v ec-re o: l)JtlC,r1...L\-M

3.·Right Angle Trigonometry

Unit 3 - Right Triangle Trigonometry - Homework

I. For each problem, calculate the answer to 4 decimal places. In the case of inverse trig, assume decimaldegrees unless otherwise stated. If a problem is impossible, state so. (answers are given on theIast pageof this packet because you need to know whether or not you are getting these correct).

# Problem Answera. sin 82° 0,910e..b. cos77° (),2'2'fq

c. tan13.6° o.2'1IGId. cos25°12' O,?o4/e. tan 225°28'49/1 I, 0 I Go rf. sin95°30" o• 99(;'1g. csc49° /,3250h. sec(-:24°) /,09'1'1. cotO.543° /05:S139J. sec45°18'9/1 /,'12/7k. cscl l °59/1 5", '2 J3(l. cot2 - o. v s r c.

# Problem Answerm. sin' .5 30-n. cos" 0.1976 7 a . " 033

0

o. arctan-J6 «t. 792..30

p. arccos(L3761) DMS !1/S7" ~H,

q. arcsin(0.7901) DMS o I '1s: 2. /1 '/ I,'I~r. tan " V10 DMS fc, :;0&, , 4, 3 ("s II

s. csc' 4.25 13;bCJ89°t. see" 0.275 ;/0'1 f/o,S'J,

u. arccot2 '2.G" 5'6 S-o"v. arcsec(3.895) DMS 75'°'1 '2 3, 9 Co ,1

w. coc'{5 DMS '2 '705/'i( '1J3 I;

x. arccsc(2.95) DMS 1'1048 'S3,7'11 II

2. Solve the following right triangles. Sides to 2 decimal places, angles in decimal degrees unless otherwisestated. Show how you got your answers.

A=62° a= 11.~~~~a) B= 2~o b= ~,3~9'1

C=90° c= 20

CJ~ e e.u:7.. ).J 2.. = c. z, v'.~ +

o I QVA= s:-e z8 a= (". \r'/og

b) B=31032' ® b= 3~ 7 <0(" t.f

C=90° c = 7.24

G >: '1o-~2 = Z @:.

0..'" a .s~ b2.1,;- = c. c.0i, '2 e

~ (po

31 32-

5"Bo '2 e'~ -::: s"""-Ac.. ®D..:. c. .ttM A

~-----'-c.(3

(3 ...::--='----:-Q----'--'

k- - ,..- Ac.. - VV.l

Js -= c. c..o:>A

A = 29.3°«I

c)B= 10.'1C=90°

a= <7. G7t;;Bb = 13.5c= IS', iBoL..f

3. Right Angie Trigonometry

qo'Z 9,3-rt 0 .'1

~~bcvnAQ~Jr'~A

- 9 - www_mastem1athmentor.com - Stu Schwartz

A= 21.75 ®d) B = 68.25°

C=90~

a= ~.q71Lj~b = 2.45 inchesc= '2...~377hl

B

~~.o.h-~2."i) c.

~ -= ~ A 0-..-= Jr~A-}..r

J.J- _ Co? A e. == 1r-c:-- 6..r.,Pr

oA= '2.6.072 if

e)B= Co\.q'2/~C=90°

a= 8b= Is----c = 17

l!l / II

A = '-llo .35 26, 10

f) B = 4 3021-/ 13 I . 8 9 '/C=90°

a = 9.25

b = 8.75c= 1'2..732.8

(Angles in Deg-Min-Sec) -

~B:: ±o,

(5~ b,,:1 ~ ®~ It = ~

-\ o, @f\=~bA

oA= 33,9S-G3

g) B = 5"6, () t.f"3 C. a

C=90°

b = 8,000 feetc = \ '1 :)Z 2, S-I 3 7

~~--x. + 8.'2. 'X. @

l ~ :C~~7-'~ =.~/ 1(-1-8:2.,.. ~

~o ~ ® r e. r-)(-\-8.'2. -=- A -x. -:.® x-;:: 7(3'~13 t- C \QJ

8.2. x' 0\- 8, Z ": ® X = ~ - 8 I-~

3. In the figure above find segment CD if LA = 2r,LB = 51 ° and AB = 8.2 miles

3. Right Angle Trigonometry

c .c®--B.2 -:::®

~ 7:-® - ® -::Q,t

c(®_®)~~'22-=- e.l\ ..L

@-®t. =- 1.! 12.9 ('(Y\ ~,

- 10 - www.mastermathmentor.com - Stu Schwartz

For each word problem, draw a picture, fill in given sides and use variables for sides and angles you don'tknow, then show equations (and/or trig functions) to find out the desired information and circle youranswer(s) being sure to use proper unit.

4. A surveyor wants to determine the horizontal distance that the top of a slope is from a tall house at thebottom of the slope. He measures the distance along the slope to be 125 feet. The angle of depression

_created from the top of the slope to the top and bottom of the house is 17°. Find the horizontal distancefrom the top of the slope to the house.

------'-'-----='t-.-J.---- .

5. What is theangle of elevation of the sun when a 6'3" man casts a 1O.5-ft shadow?):(, /.! -.I?

<, <: 0+ 12. :; (.,.Z> j-t-;

~ -= ~f)- ro, S

6. A street in San Francisco has a 20% grade - that is it rises 20 feet for every 100 feet horizontally. Findthe angle of elevation of the street?

,'//~ ~/). ~ f:4"t (;c/ -:.~

7. A kite is 120 feet high when 650 feet of string is out. What angle of elevation does the kite make with theground? ;2. 0

~r.,t):! ~

8. From a balloon 2500 ft high, a command post is seen with an angle of depression of 8°45'. How far is it.from a point on the ground below the balloon to the command post?

2:;(J 0 L 0 /-r ":~I? 'I';'2 S-D 0 ~

d ~ ~(g+ ~~")-=:::. it;. 21-2, cP.fr.

~€~~;J~__~-----------_---~¥Sl--~~.!.---1""''''------- d. ..,.1

3. Right Angle Trigonometry . - 11 - www.mastermathmentoLcom - Stu Schwartz

9. An observer on a ladder looks at a building 100 ft away, noting that the angle of elevation of the top ofthe building is 28°40' while the angle of depression of the bottom of the building is r30'. How tall isthe building? I

h -e h'f + hG ,he,. -c, ~(730/)

®lo~ ~I

-:'(D + I h = ~D() ~7.5'C>=. /3,/(,:;2 @h~h-~-- ---

-2.. := ~ (z ~ °'1-0')I I"'of,I hI ~ I be ~~ •.•~ § I ::: 53.9/7.f' (j)

10. From a plane 2 miles high, the angles of depression to two towns in line with each other are 82°55' and12°8'. How far apart are the two towns?

I 'G- ~ ~(IZ $-5')d.~J, rJz. I 4, 2

I J. I -s- =ta-.-,-=-('--8 '2..-+---=-~~=) ®- -- d~@+@

B'ZJ-S-1 I 'l °8, --~ I

,,0<,' :20>; ~~ ~1Z,*j): ::: ~(,:O") ®______ I1~ ~ ~_~11.._0.::-S_' _---"""'- I -r~ ~z.+to)k- d., >k . 42. ~

11. Jay and Sam who are both on one side of a hill are staring at the top of that 180 foot tall hill. Jay, whois nearer the hill sees the top of the hill at an angle of elevation of 72°58' while Sam, who is furtherfrom the hill sees the top of the hill at an angle of elevation of 41 °23'. How far apart are Jay and Sam?

~"IIOZ~3'"@ -&.~/ I g,b 0 I IAo

"II 62] -rz sa~\ 4+~ .p

~/ll~/80tc ;~;;@ \ /~c~~y '. ~

~ I / ..J (If /e<l ..•c~t')3' 12.°;e> / ~ +; =- 7ilJ ~

.s J ,J IJ' .: /go fJ = /80 - I~::. 19 !~o/ I~ 0\. )' r: ----J A - @ - T ® @ '.

12. Jay and Sam are still staring at the top of the hill. Jay and Sam are 75 feet apart. The angle of elevationof the top of the hill for Jay is 22° while the angle of elevation of the top of the hill for Sam is 57°.Find the elevation of the hill. UJAY A-'l.f' -rite'! £rU ...L J'T.4,zIAJ&'?

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7~@4- x@=-hl( =- h"- 7S@

@

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.L =- ts: 22' .:@7'5" fl(

I'<r- X-1.J..'= ~37°"'@'X

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13. An airplane travels at 165 mph for 2 hours in a direction of S62°W from Chicago. How far south andr">. west is the plane from Chicago?

1(" s: ""';: . '2 hA. "= 3'30 f"O'\ ~

Ir..1l.-

s~29/.~ -w

14. An airplane travels at 445 mph for 4.5 hours in a direction of 1240 from Seattle. How far south and eastis the plane from Seattle? ~ ==.A. t JL:= 4'/-'r6 t:-,:. 4. s: h Vl. l3~n.1,J4 -e 12"f '!>

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15. A ship leaves a port at 1:00 PM traveling at 13 knots directly north. Another ship leaves the same portat 2:00 PM traveling due east at 15 knots. At 9:00 PM, how far apart are the ships? Along whatheading will the northern ship have to travel to reach the eastern ship which is now stopped?

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16. A man is orienteering (traveling through forest area with only a compass). He leaves is base andtravels due north for 2.5 miles. He then travels due west for .8 miles. He then travels due south for 4miles. How far is he from his base and along what heading must he travel to reach it?

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17. A small Cessna plane leaves Naples, Florida and travels 99 miles east to Miami, Florida. It then headsdue south for 123 miles where it crashes into the ocean. If helicopter rescue is to be effected fromNaples, how far and along what heading will the helicopter have to travel?

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18. John and Ellen have a big fight and swear they will never talk to each other again. Johnleaves theairport traveling 180 miles along heading 2So. Ellen leaves the airport and travelsSdtl miles alongheading 29So. If Ellen decides she was wrong and decides to catch a flight to John that can travel at SOOmph, how long will it take her to fly into his arms? _ \ r t; __ ) Z ( _ ')7.'tJ/~ C,f IZL.s Hitrl{i Fur,) J/.,-("-d/5 O,Je:- ~ 6.s-T - \J ~I.> XI' + ~G ?,.

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Answers to Problem 1

# Problem Answera. sin 82° 0.9903b. cos7r 0.22S0c. tan13.6° 0.2419d. cos25°12' 0.9048

e. tan22S028'49" 1.0169

f. sin9S030" 0.9962g. csc49° 1.3250h. sec(-24°) 1.0946

1. cotO.543° 105.S139J. sec4S018'9" 1.4217

k. csel1°S9" S.23311. cot2 -0.4577


$- =- L -I 1'23 _ ""'/./76 II. ~ Qq - .:;T,J. c::: J qq •..•. 1231'

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~ Itfl,/70jo

5 (38. !2J?o) E )

- =su. 2)' 1!i3o 010 "IOu AJ ()·flc..~

r-;;~= '2.lf s, fill ~fi"+-2~= ~oo?

r:-~)'d;r) -x..~ ." CLl> !o ~ "J:r ~~L 1'60

,Tc; " >:tM. cr ~.,~~~s =>»: 1&',07/'2 ~:!"'l~3,tl.s1 -e, 1.."

I-P- '/(lu

~If't~r >"'PQ..J r: h.

# Problem Answerm. sin-I.5 30°n. cos-'0.1976 78.60340o. arctan-J6 67.7923°p. arccos(1.3761) DMS . impossible

q. arcsin(0.7901) DMS 52°11'41"r. tan' ViO 65.1012°s. csc' 4.25 13.6090°t. see" 0.275 impossible

u. arc cot 2 26.5651 o

v. arcsec(3.895) DMS 75°7'24"W. cocl-J5 DMS 24°S'41"x. arccsc(2.95) DMS 19°48;54"

!;o .•.+- IS'o 'L

1,1..119 Ih-tc..

c-'A.,G~ ~ AIr., .

eG WC>~I"- ~O\J,P/I G 1"'!I Dc.

- 14- www.mastermathmentoLcom - Stu Schwartz

unit 3 - right triangle trigonometry -...

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