LawofSinesCosinesArea review key

AAT Practice Test Law of Sines and Cosines
Name
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Two sides and an angle (SSA) of a triangle are given. Determine whether the given measurements produce one triangle,
two triangles/ or no triangle at all. Solve each triangle that results. Rsund lengths to the nearest tenth and angle mea$ures
to the nearest degree.
aqA,
\4{
1) B=41'
IYD*i,:,
I
f*rt llt
a=4
b=3
A)
--
78, cI
,A1 = 6!", C1 =
A/+ B = lkO
{ ttr)
_c
*:,:,it,'o ,i-ii
-- 44
t'r
t'
-
ele=2T,C=I70",c=j.7
0.1;
i
A2=119",C2=20', c2 = 0.1
,
;j,ir
At - 6L' , CI = 78", c7 = 4.5;.
A2:1I9',C2=2V,c2=1.6
C)
tto triangle
a5-r(
=
zr
B
s,rll tyj4
4 =
.rPj
!
SlnR'- 4-[rn4
A={ol
e) B=Lo2'
b-3
c4
I
3
rn{i Lirilui
.n.,*\
4!
..*,-.'.
i
\
:
''
"r,),1_;,
;!
/-
\-/
1^ ij
A)A=50",C=27',c=3A
C)A=52",C=27",c=32
D-)no triangle
.9
li,:
.ilii'i ^t n
A = 3tr
I
,rIJ
:l{t 3) a=8
i
I
!j|B.
,
it".{' /, / ,, (:
b..i6 *--tt,;.--'rF
'"/
1i:
f-,
{
^. :4
CYc: il i3."''l
B)B=60",C= 60^,c=I3.9
^
A) B = 6{}", ri
==
r)
90", c ='1.3.9
#J.B =9v,c=6a,c:Ii.e
/ \'-
i -J
' .1.*
D) no triangle
Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest
AAi
4)
B=
C:
b
degree. B
47"
:;l
100'
=20
,,4
'-.i. ,
28.9
z'--i
c=26.9 "rr-'-.-".'' , ,{,
1 c)it = 33', a = 1.4.9,
it . t .- 'l
A)A=33",a=T6.9,c
..",
i.^:
Solve the problem.
t4:
A=31',a-26.9,c:I4.9
olA=31',a=28.9,c=r6.9
B)
_.,
,-. !/'1,t1
building measures the angle to the top of the building and
findsittabeST".Thesurveyorthenmeasurestheangletothetopoftherac.liotoweronthebuildingandfincts ]-il
thatitisso',Howtallistheradiotower? 3lt'L3"f''..,t--'/55 ,L;"- r;'i,
t
B) 12,7 meters
C) 24.Lmeters
D) S.57 meters
i -A) 9.03 meters
i I .',
y.__
I
Srrl50..
ib 1
-Solve the triangle. Round lengths to the nearest tenthlnd angle mea"srires io the nearest degree.
.^u
'1I
5) A surveyor standing 55 meters from the base of
a
le
aFS
)'
ynl,
\".)C-b=1,03fn l+l
4- J
'"aFl iirJ
6)c=112" t:r.-Er+n2-Zttsiil)r,;-!t7r p{nA,_
StntiL
A:.
^
s5'"'
'-iut*
a=8
'qil'du
c'=3,5D,'f"*c't
c'15.3
f
b=11
c = 18.7,4 = 30",8 = 38"
C) c=21.6, A = 26,ts = 42'
A)
t,
'1"tilJ
/')
'p
6pt=15.8,A:2E',8=40"
D) no triangle
4gt'** ?{rz)H(nbk3"
b"l2t
v __ta-"?11
l.rtr*)dJ Lr.rdrrT SrrrA
F$y
gy-_]jl _ stnub
irsP
tF8
DB=63
a=72
12.2,
8)a:r
Z:i^
c=16
A=76',c=41"
B)
A=75",C:42
D) no tiangle
(-clSC.1'+142-tb2--tl
c-t>s c ' 71-tI*4::)k:
'-i+*
2r.1lf i1)
A)A= 28,8=59',C=93 BtnA *
sorve the
\
8ln-{?u
" -Ti'.E
z
s\v^\A-lz$rnfut
v
t\
fftlu:11,0,
\:*.,
:
338
of,fu*-**-*c b-*tO,d'lQ1trll,0
c:8
C) b
b "= \2D,K
ll
A=1tn-4
b=1.32,A=73,C:44
'\-.:r24a
cn^r..,--r/-l-_l
= c-os, f*,)
=
C
qo,zo
b= lti
B)A:21f,8:51",C=95" "-' X^,
r\
dt ,, si{i '(1:!# 3") * uo"
probrem.
t= lb
9) Two sailboats leave a harbor in the Bahamas at thei same time. The first sails at 23 mph in a direction 320". The
secon.d sails at 31 mph in a direction 190". Assuming that both boats maintain speed and heading, atler 2hours,
how far apart are the boats?
,{A}Ss.f
miter
B) 85.2
2C= 7')D-lEO'
C' f?0"
miles
cL.
C)71.8 miles
(-oZ' + 4-4,
D) 8L.5 miles
- L(r*Z)iq$ rffi
l
.*Do
L" = \bLto,4
6'fuL
J3rnph
3?o"
\zs
6-- +to
A
v(6
f,(s
vU
fl(e
f,(s
f,(7
f,(€
o(z
)(r
.!IJIOH) lIldIIlruAI
Adv Alg {'rig
W$: Area cf *i:iiq*e T:iangl*s
i*Ja:n*
lriangle are given. Sketch ani! label thc triangle, '1'1:*x. $nd the arsa *fthat
triaxgle, ineluding units" $how work. Ati final answors should bc accurale 1r: the near*sl tenths place.
'cr
eaeh problem, three pafis of,a
r, x*6m, )):
1zm,z.= g
m,
A=
-il g
i;+1.ifi-bXg-c)
1.
7t,3rn\
E''lz[urtt, lz) * g3
A={rc$,-tL\ra-rYt3-k)
-'' *.\L
J
Qr-
!
).
c
i (
\
t:
2.
z. l\. trfr"'
{i
{-
t}a,, Z" +b^
14'$ ', b"
c-r\LW
A ='tnao
a'i4'&'F''-'
b .-'t14"t
M = 55""
3.
In
A't'\
L- -.
uj-
rur
-
i"5
in,p -
15
5(m-")(=l
;-lr.g+I
:. l05,tluu"a
in
Ar ail Lflrii lS)slrl ?Do
IA
ro=ir.n;l
r
4.
1D,D-1 --
U,-15''.
v,:
F
*I05",.l;:6mi
/tas" -fp;" l(
\)
t'ffiO"
-
A
I I -, I^} {r"\i':,n
\V,Jr+'-"*
Ll1
sln
l5o
ii--t} s
,
rrt 105'
4. 58,Znnt
I
/"*'--.-,-..'e-"
I
H 'E8rr,-
n!'\
,,1*
lctrr\LiC"
VV' Y,
l'ln \5"
t0'? A ,C1
5" 3 * 9$o. C -' 3{}o, s :24
0n
3Dd
tzff
A
cm
s.l2A,"1t rmL
Adv Alg frig
WS: Area of ObXiqur:'l riangles
Narnc:
*.214' lorr'"r*'
6" Y:47u,1:25 n:lrn,z:22 mrx
SlrLZ
3rn4-7
.*--.-----",.*
i,
LP
/\
a\
-A
la
"1\ f
-- 5bz)17E)stnq"*.
4 = 174,fuffrrr]'
trnz. A*3ntl
-if
d-'\'/
Z.
. fi - ili yd
---
tt
.)
.4--
4D,Olo
n{4
q
8.
A
t\
n,*d
-'"-n
155,1
\4d'
fYJ*r"- -
I
x.
;b1
In
7'.= 57n, s ,= 7 .2 &, I
S)srri +L"
{r
*tn 5
d'L
t , t'-'
ttri$
bq I wr{,..*
r)ir a) :r{r.-r*a
,Ir,r\
j
* 3zj k,
: 9.I {1,
3
--
" #5,*{
A = ;l*
,.
s:
r(
sin'{4
*:-"--*--'-**'*
sLrl++
U
tr3 k$r, tr :
fi . #laler,
s I'n 4',1
11
v*1
lS: ?5", e; *
l8
?.
7.155,1d|t,
-o
3tn5?
,{'"
A' t {r zxq-l ) st!'t8l'4 "
I
= 1 2- il'nPjJ
*1
-t
r-*"
-- I
iA =37,4f+'
i)
\)- 41.b"
itl.u-9(t rni,v-*
120 rni,
la':
135 n:ri
[s- u)(s -v)ts-w]
frlt
/;.
i4*l't?fl +135)
\12 3
fi, \ rzz s [Bt,EXrz :)(r l)
ft : 5*J13. ?- rru ?
I
rub23(
"1
3. J f'tt,t'

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