Download 3 - Angles

December 9, 2014
Any calculator permitted on N.Y.S. Regents examinations may be used.
The word “compute” calls for an exact answer in simplest form.
3 - Angles
1. In parallelogram GRAM , m∠G = 6x + 4 and m∠R = 4x + 6. Compute x.
2. In regular pentagon BIN GO, diagonals OI and GI are drawn. Compute
m∠IN G + m∠BIO + m∠OGI.
3. In 4ABC, ∠BAC is obtuse, AD is an angle bisector of ∠BAC, m∠ABC = 60◦ , and
AC = 24. If AE ⊥ BC and BE = 2DE, compute DC.
A
B
E
D
C
December 2014
3 - Angles
1. In parallelogram GRAM , m∠G = 6x + 4 and m∠R = 4x + 6. Compute x.
SOLUTION: 17 Consecutive angles in a parallelogram add to 180◦ . Thus,
10x + 10 = 180, so x = 17.
2. In regular pentagon BIN GO, diagonals OI and GI are drawn. Compute
m∠IN G + m∠BIO + m∠OGI.
= 108. Next, m∠BIO = 180−108
= 36.
SOLUTION: 216 First, m∠IN G = 180 − 360
5
2
Last, m∠OGI = m∠GOI = 108 − 36 = 72. The desired sum is 108 + 36 + 72 = 216.
3. In 4ABC, ∠BAC is obtuse, AD is an angle bisector of ∠BAC, m∠ABC = 60◦ , and
AC = 24. If AE ⊥ BC and BE = 2DE, compute DC.
A
B
E
D
C
√
SOLUTION: 18 Note that if BE = 2x and DE = x, then AB = 4x and AE = 2x 3.
Therefore, BD = 3x. Now, applying the Angle Bisector Theorem, 3x
= DC
, so DC = 18.
4x
24
-OR1
Set up the diagram as described above. Focus on 4ADE: m∠EAD = tan−1 √ , so
2 3
◦
◦
∼
m∠EAD ≈ 16.1
.
Because
∠DAC
∠DAB,
m∠EAC
≈
62.2
.
Solving
the
equation
=
√
2x 3
◦
cos 62.2 =
yields x ≈ 3.231. Then, DC = EC − x = 24 sin(62.2◦ ) − 3.231 = 18.
24
December 3, 2013
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
3 - Angles
1. In pentagon SM U RF , three of the angles are right angles, and the other two angles are
congruent obtuse angles. Compute the degree measure of one of the obtuse angles.
2. In a quadrilateral, all angles (call them ∠1, ∠2, ∠3, and ∠4) have positive integer degree
measures, and m∠4 − m∠3 = m∠3 − m∠2 = m∠2 − m∠1 (that is, the angle measures are
in arithmetic progression). Compute the least possible degree measure of the least angle.
3. In the accompanying diagram, line T A is parallel to line OP . In convex hexagon
P OST AL, all interior angles have integer degree measures and m∠L > m∠T AL. Compute
the maximum sum of the measures of the largest and smallest angles in P OST AL.
Reminder: The diagram is not necessarily drawn to scale.
T
2x°
A
x°
L
S
3x°
O
6x°
P
December 2013
3 - Angles
1. In pentagon SM U RF , three of the angles are right angles, and the other two angles are
congruent obtuse angles. Compute the degree measure of one of the obtuse angles.
SOLUTION: 135 The sum of the angles of a pentagon is 180(5 − 2) = 540. Subtracting
out three right angles and dividing yields 540−270
= 135 degrees each.
2
2. In a quadrilateral, all angles (call them ∠1, ∠2, ∠3, and ∠4) have positive integer degree
measures, and m∠4 − m∠3 = m∠3 − m∠2 = m∠2 − m∠1 (that is, the angle measures are
in arithmetic progression). Compute the least possible degree measure of the least angle.
SOLUTION: 3 We can represent the angles as c, c + d, c + 2d, and c + 3d. The sum
4c + 6d = 360 → 2c + 3d = 180. Since c and d must both be positive integers, we consider
possibilities. If c = 1, then d = 59.333..., which is not integral. If c = 2, then d = 58.666...,
which is not integral. If c = 3, then d = 58, and the angles are 3◦ , 61◦ , 119◦ , and 177◦ . Our
answer is 3.
3. In the accompanying diagram, line T A is parallel to line OP . In convex hexagon
P OST AL, all interior angles have integer degree measures and m∠L > m∠T AL. Compute
the maximum sum of the measures of the largest and smallest angles in P OST AL.
Reminder: The diagram is not necessarily drawn to scale.
T
2x°
A
x°
L
S
3x°
O
6x°
P
SOLUTION: 205 Draw a line through S parallel to T A. Two alternate interior angle
pairs are formed, one pair measuring 2x◦ and the other pair measuring 3x◦ . Thus,
m∠S = 5x. Similarly, m∠L = 7x. The other four interior angles are supplements of the
marked angles, so they measure 180 − x, 180 − 2x, 180 − 3x, and 180 − 6x.
The largest angle must be ∠L or ∠T AL. To guarantee that ∠L is the largest, we need
7x > 180 − x → x > 22.5. Since m∠L < 180 and x is an integer, x ≤ 25. Thus, we examine
measures for x = 23, x = 24, and x = 25.
For x = 23, the six angles in the hexagon are 42◦ , 111◦ , 115◦ , 134◦ , 157◦ , and 161◦ , so the
sum of the largest and smallest is 203◦ . For the other possible values of x, the smallest
angle will be 180 − 6x and the largest will be 7x, so that as x increases by 1, the smallest
angle decreases by 6 but the largest angle increases by 7 for a net change in the desired
sum of +1. The required sum is 30 + 175 = 205.
December 4, 2012
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
3 - Angles
1. An angle is equal to the complement of the supplement of three times the angle.
Compute the measure of the angle, in degrees.
2. In the accompanying diagram, AB k CD, m∠1 = m∠2 + 15◦ , m∠3 = 73◦ . Compute
b − a.
A
.B
.
2
3
b
1
.
C
a
.
D
3. Two angles in an isosceles triangle measure (x + 5)◦ and (2x − 30)◦ . Compute the sum
of the measures of all possible vertex angles of the triangle.
December 2012
3 - Angles
1. An angle is equal to the complement of the supplement of three times the angle.
Compute the measure of the angle, in degrees.
SOLUTION: 45 We solve x = 90 − (180 − 3x) → x = 3x − 90 → x = 45.
2. In the accompanying diagram, AB k CD, m∠1 = m∠2 + 15◦ , m∠3 = 73◦ . Compute
b − a.
A
.B
.
2
3
b
1
.
C
a
.
D
SOLUTION: 78 Let m∠2 = x◦ and m∠1 = (x + 15)◦ . Then
m∠3 = 73◦ → x + 15 + x + 107 = 180 → x = 29, which means m∠1 = 44◦ and m∠2 = 29◦ ,
so b = 151. Since a is vertical to the alternate interior angle of ∠3, a = 73. Thus,
b − a = 151 − 73 = 78.
3. Two angles in an isosceles triangle measure (x + 5)◦ and (2x − 30)◦ . Compute the sum
of the measures of all possible vertex angles of the triangle.
SOLUTION: 222 If both given angles are the base angles, x + 5 = 2x − 30 → x = 35,
and then the base angles are 40◦ , so the vertex angle measures 100. If both base angles
measure (x + 5)◦ , 2(x + 5) + 2x − 30 = 180 → x = 50, and the vertex angle measures
2(50) − 30 = 70. If both base angles measure (2x − 30)◦ , 2(2x − 30) + x + 5 = 180 → x = 47,
and the vertex angle measures 47 + 5 = 52. The desired sum is 100 + 70 + 52 = 222.
December 13, 2011
Calculators are not permitted on this contest.
The word “compute” calls for an exact answer in simplest form.
6 - Angles
1. From the set {acute, right, obtuse, straight}, choose the word that best describes the
supplement of the complement of an acute angle.
2. In 4ABC, m∠A = 4x + y, m∠B = 8x − y2 , m∠C = 7x + 4y, and an exterior angle at B
measures 10x + 10y − 1. Compute the ordered pair (x, y), given that AB = AC.
3. In right triangle ABC, the angles ∠GF I, ∠HF I, ∠IHK, ∠IKH, and ∠JBK have
measures as indicated in terms of x and y. Also, m∠GDF = 36◦ and five perpendiculars to
BC are marked. If m∠CAD = a and m∠EDG = b, compute a + b, in degrees.
A
a D
b 36°°
F
y x-3
H
y+10
J
x+20
C
E
G
I
x
K
B
December 2011
6 - Angles
1. From the set {acute, right, obtuse, straight}, choose the word that best describes the
supplement of the complement of an acute angle.
SOLUTION: obtuse The measure of the supplement of the complement of an acute angle has
measure 180 − (90 − x) = 90 + x. Since x lies in the interval 0 < x < 90, then we know that
90 < 90 + x < 180, so our answer is obtuse.
2. In 4ABC, m∠A = 4x + y, m∠B = 8x − y2 , m∠C = 7x + 4y, and an exterior angle at B
measures 10x + 10y − 1. Compute the ordered pair (x, y), given that AB = AC.
SOLUTION: (9, 2) Since base angles of an isosceles triangle are congruent, we have
7x + 4y = 8x − y2 → x = 92 y. Since the measure of an exterior triangle equals the sum of the two
remote interior angles, we have 10x + 10y + 1 = 4x + y + 7x + 4y → x = 5y − 1. Substituting, we
have 5y − 1 = 92 y → y = 2, and so x = 9. We write the answer in the desired form: (9, 2).
3. In right triangle ABC, the angles ∠GF I, ∠HF I, ∠IHK, ∠IKH, and ∠JBK have measures
as indicated in terms of x and y. Also, m∠GDF = 36◦ and five perpendiculars to BC are
marked. If m∠CAD = a and m∠EDG = b, compute a + b, in degrees.
A
a D
b 36°°
F
y x-3
H
y+10
J
x+20
C
E
G
I
x
B
K
SOLUTION: 78 In 4HIK, x + y = 60. Applying the fact that the measure of an exterior
angle equals the sum of the measures of the two remote interior angles to 4HKB forces
m∠KHB = 20. ∠GF H and ∠IHJ are corresponding angles of parallel lines, so
x + (y − 3) = (y + 10) + 20, which means x = 33, so a = 57 and y = 33 and m∠DF G = 123. As
alternate interior angles of parallel lines, m∠DGF = m∠EDG = b. Therefore, in 4DF G,
b = 180 − (36 + 123) = 21. Our answer is a + b = 57 + 21 = 78.
Authors: Michael Curry - [email protected] – George Reuter - [email protected]
www.mathmeets.com
DECEMBER 14, 2010
NO CALCULATORS ON THESE TOPICS
#6
ANGLES
ANSWERS:
1.
2.
3.
1. Find the number of degrees in the sum of x and y given:
m∠ANB = 32°, m∠ANF = (5x + y)°,
m∠BNC = 47°, m∠FND = (4x − y)°
2. An exterior angle of a regular polygon is
How many sides does the polygon have?
1
35
of an interior angle.
3. CD EF and AB is a transversal intersecting
CD and EF at points M and N respectively.
P is a point between the parallel lines such that
m∠NMP = 3 m∠PMD and m∠MNP = 4 m∠PNF.
If m∠AMD = (7x − 40)° and m∠MNF = (5x)°, find m∠P.
DECEMBER 15, 2009
NO CALCULATORS ON THESE TOPICS
#6
ANGLES
ANSWERS:
1.
14,040
2.
130
3.
12 + 5 3
1. One exterior angle of a regular polygon measures 4.5˚.
What is the sum, in degrees, of the interior angles of this polygon?
2. In ∆ABC , AB = BC , m∠A = 50. If point P lies in the interior of ∆ABC
such that ∠PBC ≅ ∠PCA, find the number of degrees in the m∠BPC.
3. The area of the right trapezoid shown is 120 cm2.
Compute the number of cm in the length of the
longest side of the trapezoid.
DECEMBER 9, 2008
NO CALCULATORS ON THESE TOPICS
#6
ANGLES
ANSWERS:
1.
112˚
2.
45
3.
50˚
1. Given: BE DF , m∠GDF = 24 , m∠ABE = 88.
Find: m∠BCD.
E
A
B
F
C
G
D
2. Two angles are such that the ratio of their complements is 4 to 3, while the ratio of their
supplements is 10 to 9. Find the number of degrees in the larger angle.
A
3. In ∆ABC, m A = 80˚ and BD and DC are bisectors
of exterior angles B and C respectively. Find m D.
B
C
D
NI@NROE C@LTNT'Y
N{AT]I{ [,]EAGU]E
DECEMBER9, 2OO8
NO CALCALATORS OAITHESE TOPICS
#6
ANGLES
ANSWERS:
1.
112"
2.
45
50"
a
1
1. Given:BE ll DF,nIGDF = 24',mlABE = 88".
Find: mtBCD.
l8D
,to
- ?B tii
{gf,il-
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bcD Fa ; 5+o'
+4tr
":*,:g'*'Il
".^nr'
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.1
t76,
u \r
2. Two anglesare suchthat the ratio of their complehen*je4 to 3, while the ratio of their
supplements
is 10 to 9. Find the numberof degreesin the larger
Y angle.
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J.f
' l l u- * = t?q
q
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2.' ?0-JX-3 uo -* 1,
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--) ( +Y-i* ' t*)
lg*;g *,,' te
T
-Bn.* lq
l(p Lt ) : qf s ls r ) O - l0i
toy -Ax t l8D
3. In AABC, mlA= 80" and BD andDC nebisectors
of exteriorangles
B andC respectivety.
Findm.r,D.
aa -ba , kr - (a-. -L/
# s / r + . { " * 0 **a-tfcc&r*c
w'A'r'
;iC,
: lf u. . t 1r t *r =) : J " Y
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ly - :
t
-11=-qjt
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v\\trIre ..*rt-c-r i-t.4- 9{n
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U
NAONRO]ECO{JN]TY
NAATFil ]L]EAGU[]E
DECEMBER
11,2OO7
lYO CALCULATORSO]YTHIS TOPIC
#6
ANGLES
ANSWERS:
54
400. 4go
3
2400
t . In AACI, mlA: 2x,mlC: 4x - 12,mlI:3x - 6,
a n d m l AMG:x*y.
i$ s t*.6
Find thevalueofy so thatMG is parallelto CL
x+Y:
{ x-l?-
1 = 3x-l?-
J = g{eu}-ft-
?# + {x-ll t 3X-LE lfiu
I x -l?: l9*s
4 x. \ 9t
N= 2Z
Findall possiblevaluesfor the mlB,
expressiqgyour answer(s)in degrees.
6
y=t8o
arll€^ xrX"\
f3.\Uxtl*'"fj!\
'fSJt-ffits/
(.x"tF)
--
3 . Eachangleof an equilateraltriangleis divided
into 4 equalanglesas shown. Find the sumof the
measures
of IADF, IAE}J, IFGC, and IAFD,
expressingyour answerin degrees.
b0 1{:15
{ roF.: lq:-g
4 A F''g*o' Go
d FGL-- 45
r AF$; 3o
\tr
)(i
)ffif
of c*
0f
hr&I"its€
ry
f69sit\e
so[u-*r6i.-+. Tk p,_gte,ts
NAO}$R'OECO{JT\$TY
}\4A1['F{]I,EAG
DECEMBER12,2006
NO CALCULATORSON THIS TOPIC
#6 ,
,
ANGLES
ANSWERS:
I
L300
150
3
1290
fNote: Expresseachanswerin degrees]
1 . F i n dm l L
L x + )Z = loo
X+- ;t'- 5 O
lh
--<o
t30
wICDE = 90o,
2. Glventwo parallellines.If nIABC = 3x -5,mlBCD = 110o,
f
andnIDEF: x 5. calculatethevalueof x.
20 + X+5=3ur-5
b+ f = 3*-s
3a=2{
3. Three lines meet at a point forming 6 angles,three of which are (x + 10)", (3x -25)o,
and (x + 15)o. If none of theseanglesare equal,find the sum of the smallest and largest
of the three.
57,.ro =lto
ar= tffi
\= bL
X{ - ts = *6 ' r
I
3F->a - 83 -)
XttS= 5l
-S Jrr1
=