I

I

l. True or False

Fifth Long Exam

{a} Theexpression i2o0e q 12010 a 12011 .rn be simplified into i.(b) The law of sines and the law of cosines also hold true for right triangles.(c) The rectangular coordinates (n,-n) converted to the polar coordinates ls (frn, I).(d) The equation cos-l(cos x) = x is an identity.(e ) c i sn -1=0(f) The standard polar form of -3y'3 * 3i is 6 cis f.(g) The simplified cartesian torm otffiis oi.

ll. Solve for all possible values of r eC.

(a) sinx = cos x ,x in radians, r €R

(b) s in6r * coss x = 6,0 1 xS 180"

(cl x7 - 2x3 + ZtEix3 = o

Simplify and express in Cartesian form

( /5+ i)" -(z+zi \E)6

(3 cis zoo)2(z cis 1oo.)18 cis 50"

(+ cls rss")2(\6+i)3(2 cds 48o)s

lV. A right triangle has sides of length 9 and 11. What are the possible lengths of the third side?

V. Solve the following completely.

(a) Find the 4th roots of z = 3 - 3i. Express answer in polar form.

(b) Express (-i * 9)t ,rr atgebraic form a * bi.

(c) Given zt = 2 - 2i and zz = fi cis(60"). Find the value of (z)(223) in rectangular form.

(d) Given z = -4 * 4i. Express in polar form for all 9, wheregis an argument of z.

ilt.

(a)

(b)

(c)

f

Vl. Problem Solving

(a) When the angle of elevation of the sun is 60", a telephone pole that is tilted atangle 20o directly away from the sun casts a shadow 30 feet long on level ground.Find the length of the pole.

(b) In a children's park, a 28-feet slide makes an angle of 30o with the ground. lts topis reached by a ladder inclined 45o to the ground. Find:i. the length of the.ladderii. the distance of the top of the slide to the ground.

(c) At some distance from the base of a giant tree, it is found that the angle ofelevation of the giant tree is 30o. After walking a distance of 100m closer to thegiant tree, it is found that the angle of elevation of the same top is now 60o. Howgigantic is the tree?

Fifth Long Exam Answer Keyl. True or False(a) False (b) True (c) False (d) False

11. (alsinx=cosx, x in rad, xelRsinx- -=1cosx

tanx = 1Possible x-l+ Z*n

-r 54 * zkn

4For all integers k

(b) srn6r + cos6x = 5, 0o ( x ( 180oSince - ! 3 stnx, cosx 3 1, then 0

:. sin6x + cos6x (-2,No solution to sin6x+cos5x=5

lll. Simplifying(a)

(b)

(V3+ t)tz -Q+zi\fr6

: ,, (f* l\rrz - r' lL . t/3\=tr\Z*i lr ' -t*\t* , )- (2cds30")tz - (4cis6}")6= 4096cis360o - 4096cis360"=E(zciszo")z (zcis1oo")

18cisS0"(9cis40")(Zcfs100")

18crs50o18cis140'=ff i=cds9Oo=El

(+cisrgs")z (y'5+i)3(2cfs48")s

(7 6 c is27 0") (2 cds3 0") 3

32cis240"= 4cisL20"

=-(-:.f)=n+zfiil

Case 1: 11 is the hypotenusegz +b2 = ! ! ?

b2=40

b=V46' - ZtltoCase 2: b (3'd side) is the hypotenuse92+LLz-b2b2 = 202

b = Zffi or '1202

lu

{e) False (f) True (e) True

(cl x7 - 2x3 + 2rl1ix3 = o* t ( r r -z+z^6t) -s

x=0 ,x4-2+z t6 t=0x4 =2-x l \ r

xa=4/1-€ ' \\z-T')xa = 4cis300o

'x=Q,c is (75 '+90"k)

fo rk=0,L ,2 ,3:. x = 0, cis75o,cist65o, cisZ55o, cds345o

V. (a)z=3-3i

^ n / J1 J t . \ 5n= 3"12 (..; - ii) = 3"l2cis71 - S l t . .

2a -_ (3,,lZcds_n )a5rr ^ ,

- (t'll)ictr'*rtno'+k = 0 ,L ,2 ,3

:. the 4throots of z arei- r 5n - 73n

(3,'12)zcis U,

(3,f 2)' cis r-,

(t,/T)+ c i s'f , {t Ar' o'#

, , . - r . € . zntD , t+V r=cLSV

, -1 .V3 . . " . Z f t - o . L6n(T * T i)' = (cis 5)o

= cis -3-I...................'.:-l

l -1 v3. l=lT-T'l

(C)zz3 = (y'tcrs60')3

= 3r6cis180" = -3y'3r

21223 = (2-Zi)(4\Ei)

= l-6v3 - 6V3r I{d) -4+4i=4(-1+i}

(c)

lv.

v1.

= 4,tze+**o12 \ tZ '

Probfem Solving(a) Let h=length of pole

, i

= E]k;l3s''l

100Let y=6g1961of the tree

tan60"=Zx

y=x(tan50"))y=x€

[3nJQ"3410O+x

y=(100+x)(tan30')

=(roo+x)(f)

y=(loo+x)(€J=xy'3(loolxx\f3) = 3rlix1OO\E+r6x=3rf3x10o\tr = 2x#

I

1200

.,:g_0:.II

30ft0=90o -20o=70

a = 180o -70 - 30o = 80o

30fr h- - - = -sin8Oo sin3Oo

. 3 0 t o - 1 5 c ,n =;;m;'{c - f f i I t

ii. In a 30-60-904 the length of the

side opposite the 30o angle is f the

length of the hypotenuse. .'.

*=]tza)ffi

in an isosceles triangle, the fengths of^ I t

the legs are j times the length of the

. { 2 2 x 2 A. ' . X = - - Y J } / = ; r = - -. \ " f r -hypotenuse

Wre