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.
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