Chapter 12 Exercises page 140 1- a 18 23 30 37 55 60

Chapter 12
Exercises page 140
1a
sin a
cos a
tan a
18
0.30
0.95
0.32
23
0.39
0.92
0.42
30
0.50
0.87
0.58
37
0.60
0.80
0.75
55
0.82
0.57
1.43
60
0.87
0.50
1.73
2-
CP
CA
AP
sin C
sin P
cos C
cos P
tan C
tan P
Case 1
10
8.66
5
0.5
0.86
0.86
0.5
0.57
1.73
rules used:
sin C = opp/hyp = AP/CP
sin P = opp/hyp = AC/CP
CP2 = AP2 + AC2
cos C = adj/hyp = AC/CP
cos P = adj/hyp = AP/CP
tan C = opp/adj = AP/AC
tan P = adj/hyp = AC/AP
Case 2
10.44
8
6.72
0.64
0.766
0.766
0.64
0.84
1.19
Case 3
5
4
3
0.6
0.8
0.8
0.6
0.75
1.33
Case 4
15
12
9
0.6
0.8
0.8
0.6
3/4
1.33
Case 5
15
9
12
0.8
0.6
0.6
4/5
1.33
0.75
3a- (a)
using Pythagoras theorem:
AC2 + CB2 = AB2
AC2 = AB2 – CB2 = 25 – 16 = 9
AC = √9 = 3
then:
sin x = opp/hyp = AC/AB = 3/5
cos x = adj/hyp = BC/AB = 4/5
tan x = opp/adj = AC/BC = 3/4
(b)
using Pythagoras theorem:
AB2 + AC2 = BC2
AB2 = BC2 – AC2 = (5√2)2 – 52 = 50 – 25 = 25
AB = √25 = 5
then:
sin x = opp/hyp = AB/BC = 5/5√2 = √2/2
cos x = adj/hyp = AC/BC = 5/5√2 = √2/2
tan x = opp/adj = AB/AC = 5/5 = 1
(c)
using Pythagoras theorem:
AC2 = AB2 + BC2 = 32 + 62 = 9 + 36 = 45
AC = √45 = 3√5
then:
sin x = opp/hyp = AB/AC = 3/3√5 = √5/5
cos x = adj/hyp = BC/AC = 6/3√5 = 2√5/5
tan x = opp/adj = AB/BC = 3/6 = 1/2
(d)
using Pythagoras theorem:
AC2 = AB2 + BC2 = 32 + 152 = 234
AC = √234 = 3√26
then:
sin x = opp/hyp = AB/AC = 3/3√26 = √26/26
cos x = adj/hyp = BC/AC = 15/3√26 = 5√26/26
tan x = opp/adj = AB/BC = 3/15 = 1/5
b- (a)
x = sin-1(sin x) = sin-1(1/2) = 30
BAC = 180 – (ABC + BCA)
= 180 – (90 + 30)
= 60
(b)
x = sin-1(sin x) = sin-1(√2/2) = 45
ABC = 180 – (BAC + BCA)
= 180 – (90 + 45)
= 45
(c)
x = sin-1(sin x) = sin-1(√5/5) = 27
BAC = 180 – (ABC + BCA)
= 180 – (90 + 27)
= 63
(d)
x = sin-1(sin x) = sin-1(√26/26) = 11
BAC = 180 – (ABC + BCA)
= 180 – (90 + 11)
= 79
4a- sin D = opp/hyp = OG/DG = 4/DG = 4/√41 = 4√41/41
we have:
DG2 = OG2 + OD2
= 42 + 52 = 16 + 25 = 41
→DG = √41
cos D = adj/hyp = OD/DG = 5/√41
tan D = sin D / cos D = 4√41 / 5√41 = 4/5
b- DG2 = OG2 + OD2
= 42 + 22 = 16 + 4 = 20
→DG = √20 = 2√5
sin D = opp/hyp = OG/DG = 4/2√5 = 2√5/5
cos D = adj/hyp = OD/DG = 2/2√5 = √5/5
tan D = sin D / cos D = 2√5 / √5 = 2
5APR= 90
SCA = 90
tan A = opp/adj
in triangle ACS: tan A = CS/AS
in triangle APS: tan A = PS/PA
in triangle APR: tan A = PR/AR
in triangle ACN: tan A = CN/AC
→ tan A = CS/AS = PS/PA = PR/AR = CN/AC
6a- 10/100 represents the sin of the angle
sin x = opp/hyp = 10/100
b- sin x = opp/hyp
10/100 = 60/d
d = 600 meter
7a- a = tan α
2 = tan α → α = 63.4
b- a = - tan α
3 = tan α → α = 71.6
c- a = tan α
1.5 = tan α → α = 56.3
d- a = tan α
2.3 = tan α → α = 66.5
e- a = tan α
0.25 = tan α → α = 14
8a- A(2, 3) and B(1, -2)
b- tan α = slope of (AB)
a(AB) = (yB – yA) / (xB – xA) = (-2 -3)/(1 – 2) = -5/-1 = 5
→ tan α = 5
c- α = yan-1 (tan α) = tan-1 (5)
d- equation of (AB) is of the form : y = ax + b
→ y = 5x + b
A belongs to (AB) → its coordinates verify the equation of (AB)
→ yA = 5xA + b
3 = 5(2) + b
3 = 10 + b
→ b = 3 – 10 = -7
then: y = 5x – 7
9a- P(5, 2)
α = 30
(d): y = ax + b
a = tan α = tan 30 = 0.57
P belongs to (d) → yP = 0.57xP + b
2 = 0.57(5) + b
b = -0.85
therefore, (d): y = 0.57x – 0.85
b- P(-1, 3)
α = 40
(d): y = ax + b
a = tan α = tan 40 = 0.83
P belongs to (d) → yP = 0.83xP + b
3 = 0.83(-1) + b
b = 3.83
therefore, (d): y = 0.83x + 3.83
c- P(0.5, -2)
α = 15
(d): y = ax + b
a = tan α = tan 15 = 0.26
P belongs to (d) → yP = 0.26xP + b
-2 = 0.26(0.5) + b
b = -2.13
therefore, (d): y = 0.26x – 2.13
d- P(2, 4)
α = 60
(d): y = ax + b
a = tan α = tan 60 = 1.73
P belongs to (d) → yP = 1.73xP + b
4 = 1.73(2) + b
b = 0.54
therefore, (d): y = 1.73x + 0.54
11a- A(2,3)
B(1,4)
a(AB) = (yB – yA) / (xB – xA) = (4 -3)/(1-2) = -1
a = -tan α → α = 45
b- A(0,2)
B(3,-1)
a(AB) = (yB – yA) / (xB – xA) = (-1 -2)/(3-0) = -1
a = -tan α → α = 45
c- A(3,6)
B(4,5)
a(AB) = (yB – yA) / (xB – xA) = (5 -6)/(4-3) = -1
a = -tan α → α = 45
12a- (OA): y = ax (passing through the origin)
a = tan α = tan 30 = 0.57
→ (OA): y = 0.57x
b- A belongs to (OA) → yA = 0.57xA (1)
OA = 5
→ √[(xA – xO)2 + (yA – yO)2] = 5
√(xA2 +yA2) = 5
xA2 + yA2 = 25 (2)
substitute first equation in the second
→ xA2 +(0.57xA)2 = 25
xA2 + 0.3249xA2 = 25
xA2 = 18.87 → xA = √18.87 = 4.34
→yA= 0.57xA = 0.57(4.34)= 2.47
Therefore, A(3.34, 2.47)
13distance = x
tan 40 = opp/adj = 69.2/x
→ x = 69.2/tan40 = 69.2/0.83 = 83.37 m
Chapter 12
Problems page 143
1-
a- APL = 180 – (PLA + LAP)
= 180 – (30 + 90)
= 60
(sum of angles in triangle = 180)
triangle APL is right, and [AM] is the median relative to the hypotenuse [PL]
then, AM = MP = ML (property)
→ triangle PAM is isosceles having an angle APM = 60
so, triangle PAM is equilateral.
→MPA = PMA = MAP = 60
triangle MAL is isosceles (proved)
MLA = MAL = 30
AML = 180 – AMP
= 180 – 60
= 120
(AML and AMP are supplementary angles)
b- we have LP = 2MP since M is the midpoint of [LP]
but MP = AP (MAP is equilateral triangle proved)
therefore, LP = 2AP
c- given LP = a
We have triangle PAL is semi-equilateral (proved)
then: side facing 30 = hyp/2
→
AP = LP/2 = a/2
and side facing 60 = hyp√3/2
→
LA = LP√3/2 = a√3/2
d- sin 30 = opp/hyp = AP/LP = a/2 / a = 1/2
cos 30 = adj/hyp = LA/LP = a√3/2 / a = √3/2
tan 30 = sin 30 / cos 30 = 1/2 / √3/2 = √3/3
e- sin 60 = opp/hyp = LA/LP = a√3/2 / a = √3/2
cos 60 = adj/hyp = AP/LP = a/2 / a = 1/2
tan 60 = sin 60 / cos 60 = √3/2 / 1/2 = √3
2a- ROI = 90 (given)
ROI is isosceles → OIR = ORI
sum of angles = 180
ROI + ORI + OIR = 180
90 + 2ORI = 180
→ ORI = (180 – 90)/2 = 90/2 = 45
OIR = 45
b- using Pythagoras theorem:
IR2 = IO2 + OR2
a2 = IO2 + IO2
(IO = OR isosceles)
2
2
2
2
a = 2IO → IO = a /2
→ IO = OR = a/√2 = a√2/2
c- sin 45 = opp/hyp = OR/IR = a√2/2 /a = √2/2
cos 45 = adj/hyp = IO/IR = a√2/2 /a = √2/2
tan 45 = sin 45 / cos 45 = √2/2 / √2/2 = 1
3a- R.T.P: cos4x – sin4x = cos2x – sin2x
Proof: cos4x – sin4x = (cos2x)2 – (sin2x)2
= (cos2x – sin2x)(cos2x + sin2x)
= cos2x – sin2x since cos2x + sin2x = 1 (property)
b- R.T.P: cos2x – sin2x = 1 – 2sin2x
Proof: we have: cos2x + sin2x = 1
→ cos2x = 1 – sin2x
Then: cos2x – sin2x = (1 – sin2x) – sin2x
= 1 – 2sin2x
c- R.T.P: cos2x – sin2x = 2cos2x – 1
Proof: we have cos2x + sin2x = 1
→ sin2x = 1 – cos2x
Then: cos2x – sin2x = cos2x - (1 – cos2x)
= 2cos2x – 1
d- R.T.P: 1 + tan2x = 1/cos2x
Proof: 1 + tan2x = 1 + (sinx/cosx)2
= 1 + sin2x/cos2x
= (cos2x + sin2x)/cos2x
= 1/cos2x
e- R.T.P: (1 + tan2x)/tan2x = 1/sin2x
Proof: (1 + tan2x)/tan2x = ( 1 + sin2x/cos2x) / (sin2x/cos2x)
= (cos2x + sin2x)/cos2x / (sin2x/cos2x)
= (1/cos2x) / (sin2x/cos2x)
= cos2x / (sin2xcos2x)
= 1/sin2x
4a- in triangle ABH, we have AHB = 90
→ sinẞ = opp/hyp = AH/AB
→ AH = AB.sinẞ = c.sinẞ
(AH altitude)
in triangle ACH, we have AHC = 90
→ sinϒ = opp/hyp = AH/AC
→ AH = AC.sinϒ = b.sinϒ
(AH altitude)
b- Let [BI] be the altitude issued from B
In triangle BIA, we have BIA = 90
→ sinα = opp/hyp = BI/AB
→ BI = AB.sinα = c.sinα
In triangle BIC, we have BIC = 90
→ sinϒ = opp/hyp = BI/BC
→ BI = BC.sinϒ = a.sinϒ
From the two triangles, we have:
BI = BI
c.sinα = a.sinϒ
c- From part (a), we have: c.sinẞ = b.sinϒ → sinẞ/b = sinϒ/c
From part (b), we have: c.sinα = a.sinϒ → sinα/a = sinϒ/c
then: sinα/a = sinẞ/b = sinϒ/c
5
Given: OA = 1cm
R.T.F: AB, BC, CD, DE and EM
We have:
tan 30 = opp/adj = AB/OA = AB/1 = AB
→ AB = √3/3
sin 30 = opp/hyp = AB/OB
→ OB = AB/sin30 = √3/3 / 1/2 = 2√3/3
In triangle OBC:
tan 30 = opp/adj = BC/OB
BC = tan 30 x OB = √3/3 x 2√3/3 = 2/3
→ BC = 2/3
In triangle OBC, we have:
sin 30 = opp/hyp = BC/OC
→ OC = BC/sin 30 = 2/3 / 1/2 = 4/3
In triangle ODC, we have:
tan 30 = opp/adj = DC/OC
→ DC = tan 30 x OC = √3/3 x 4/3 = 4√3/9
→ DC = 4√3/9
sin 30 = opp/hyp = DC/OD
→ OD = DC/sin30 = 4√3/9 / 1/2 = 8√3/9
In triangle ODE, we have:
tan 30 = opp/adj = DE/OD
→DE = tan 30 x OD = √3/3 x 8√3/9 = 24/27
→ DE = 24/27
sin 30 = opp/hyp = DE/OE
→ OE = DE/sin30 = 24/27 / 1/2 = 16/9
In triangle OME, we have:
tan 30 = opp/adj = EM/OE
→EM = tan 30 x OE = √3/3 x 16/9 = 16√3/27
→ EM = 16√3/27
path traversed from A to M:
AB + BC + CD + DE + EM
= √3/3 + 2/3 + 4√3/9 + 24/27 + 16√3/27
= (37√3 + 42) / 27
6a- The two triangles AHC and ABC are similar since angles CHA = CAB = 90 (given)
and C is a common angle.
So, ABC = α = CAH
ACH = ẞ = BAH
b- In triangle ABH: cosα = BH/AB
In triangle ABC: cosα = AB/BC
→ BH/AB = AB/BC
→ AB2 = BH x BC
c- In triangle ABH: tanα = AH/BH
In triangle AHC: tanα = HC/AH
→AH/BH = HC/AH
→ AH2 = HB x HC
7-
a- OM = OA
(radii in the same circle)
→ OMA is an isosceles triangle at O
but [MH] is a median and height at the same time
→ OMA is isosceles at M also
→ OM = MA = OA
Therefore, OMA is an equilateral triangle.
b- OHM is a right triangle
→ OM2 = OH2 + MH2
62 = 32 + MH2 → MH2 = 36 – 9 = 27
→ MH = √27 = 3√3 = 5.196 cm
c- AMB = 90
(inscribed angle facing diameter)
and MAB = 60 (triangle MAO is equilateral)
→ triangle AMB is semi-equilateral.
In a semi-equilateral triangle, side facing 60 = hyp√3/2
→ MB = AB√3/2 = 12√3/2 = 6√3 = 10.392 cm
d- sin ABM = opp/hyp = AM/AB = 6/12 = 1/2 = 0.5
ABM = sin-1(0.5) = 30
Second method:
In triangle ABM, we have:
AMB + ABM + BAM = 180
90 + ABM + 60 = 180
ABM = 180 – (90 + 60) = 30
e- Triangle ONB is semi-equilateral
(BON = 90 and OBN = 30)
→ side facing 60 = hyp√3/2
OB = NB√3/2
NB = 2OB/√3 = 2(6)/√3 = 12/√3 = 6.928 cm
→ side facing 30 = hyp/2
ON = NB/2 = 6.928/2 = 3.464 cm
8-
a- tan A = opp/adj = BC/AB
→ BC = tan A x AB
= 0.375 x 8
= 3cm
b- In right triangle ABC:
AC2 = AB2 + BC2
= 8 2 + 32
= 73
→ AC = √73 = 8.544 cm
c- Vectors AF = BC
(given)
→ (AF) // (BC)
and [AF] = [BC]
→ AFBC is a parallelogram.
but ABC = 90
Therefore, AFBC is a rectangle.
9a-
b- In triangle ABC, we have: ABC = 120
ABC + BAC + ACB = 180
but ACB = BAC (isosceles triangle)
→ 120 + 2ACB = 180
2ACB = 180 – 120
ACB = 30
In triangle HBC, we have:
HBC = 180 – (BHC + HCB)
= 180 – (90 +30)
= 60
c- Triangle HCB is semi-equilateral
sin 30 = BH/BC → BH = 0.5 x 5 = 2.5 cm
d- In triangle BCD, we have:
BC = BD
(radii in the same circle)
DBC = 180 – ABC
(supplementary angles)
= 180 – 120 = 60
→ BCD is equilateral triangle
→ BCD = 60
→ HCD = HCB + BCD = 30 + 60 = 90
Therefore, (BH) // (CD)
(2 lines perpendicular to the same line are parallel)
DC = 5cm
10Aa- 0 ≤ x ≤ 90
b- CE + ED = CD
→ CE = CD – ED
= 90 – x
Ba- tan DEB = opp/adj = DB/ED = 35/x
b- tan CEN = opp/adj = CN/CE = 25/(90 – x)
Ca- 35(90 – x) = 25x
3150 – 35x = 25x
3150 = 25x + 35x
3150 = 60x → x = 52.5
b- tan CEN = 25/(90 – x) = 25/(90 – 52.5) = 25/37.5 = 0.666
→ CEN = tan-1(0.666) = 33.66
tan DEB = 35/x = 35/52.5 = 0.666
→DEB = tan-1(0.666) = 33.66
11a- Right triangle MNP:
→ NP2 = MP2 + MN2
52 = MP2 + 42
MP2 = 25 – 16 = 9
MP = √9 = 3 m
b- sin MPN = opp/hyp = MN/NP = 4/5
→ MPN = 53.13
c- NA = NM – AM
=4–1=3m
NB = NP – BP
= 5 – 1.25 = 3.75 m
NA/NM = 3/4 = 0.75
NB/NP = 3.75/5 = 0.75
→ NA/NM = NB/NP
→ (AB) // (MP)
(inverse of Thales Theorem)
12a-
b- tan ACD = opp/adj = AD/DC = 5.4/7.2 = 0.75
→ ACD = 36.9 = 37
c- tan CAB = opp/adj = CB/AB = 5.4/7.2 = 0.75
→ CAB = 37 = ACD
d- Let H be the midpoint of [AC]
In triangle ECH, we have:
tan ECH = opp/adj = EH/HC
In triangle EAH, we have:
tan EAH = opp/adj = EH/AH
but AH = HC (given)
So, tan ECH = tan EAH = EH/HC = EH/AH
→ ECH = EAH
Therefore, ACE is an isosceles triangle.