THREE DIMENSIONAL TRIGONOMETRY - Angelfire

THREE DIMENSIONAL TRIGONOMETRYLecturer: Mrs Wong Lai Yong1Sine rule and cosine ruleASine rule :abc sin A sin B sin CCosine rule :OR :22cb2a b c - 2bc cos Ab2 a2 c2 - 2ac cos Bc2 a2 b2 - 2ab cos Ccos A b2 c 2 a22bccos B a2 c 2 b22accos C a2 b2 c 22abBaCExample 1: N2002/1/9(i)In triangle ABC, angle B θ, angle C θ α, AB 2 and AC 1. Show that tan θ Solution:sin(θ α )22 sin θ sin θ cos α cos θ sin α2 tan θ tan θ cos α sin αtan θ (2 - cos α) sin αsin αtan θ 2 cos αsin α.2 cos αAsin θ 21θθ αBCExample 2: J95/1/4The diagram shows triangle OAB in which OA 2 units, OB 1 unit and angle OAB 30o.The point C is the foot of the perpendicular from O to the line AB produced.(a) Find angle ABO, and show that AB 2 sin 15o.(b) Hence, by using triangle OAC to find AC, show that the exact value of sin 15o is6 2.O4Solution:(a) 21Let angle ABO θA2003Three-Dimensional Trigonometry LYWong30oBC1

1sin 30 o 2sin θsin θ 2 sin 30 o 22 θ 135 oangle AOB 180o - 135o - 30o 15oAB2 osin θsin15AB (b)22 2 sin15 o 2 sin15 ocos 30 o AC2AC 2 x ( 3/2) 6/2BC AC - AB 6/2 - 2 sin 15oBut cos OBC BC/1 BC cos 45o 2/2Hence, 2/2 6/2 - 2 sin 15o6 22 sin 15o 26 2 sin 15o 42 Angle between two lines2.1 Lines in common planeThese are either parallel, non-parallel or intersectingβL1αL2Angle between L1 and L2 is α or β.BαβACAngle between AC and AB is β.Angle between CA and AB is α.2003Three-Dimensional Trigonometry LYWong2

2.2Lines lying in different planesThese are non-parallel and non-intersecting. They are called skew lines.ABαDCEFHGAG and EF is a pair of skew lines.DH and AB is also a pair of slew lines.Definition:Angle between two skew lines is the angle between two straight lines drawn parallel to the twoskew lines through any point in space.Angle between AG and EF Angle between AG and AB α (since AB is parallel to EF)Angle between DH and AB Angle AE and AB 90oExample 3:Consider the box given below. Find(i) angle between BC and BG(ii) angle between AD and AG(iii) angle between DG and EF.ASolution:(i)tan α ¾α 36.87o(ii)DG2 32 122DG 12.35tan β 12.35/4β 72.08o412βαCD3BFEγH12G(iii) Angle between DG and EF Angle between DG and HG ( since HG parallel to EF) γtan γ 3/12γ 14.04o2003Three-Dimensional Trigonometry LYWong3

3Angles between a line and a planeQRαπPLine QP meets plane π at P. Drop a perpendicular from Q onto π. Call this QR. Then, RP isthe projection of QP onto π.Definition:Angle between a line and a plane Angle between a line and its projection onto the planeTherefore, angle between QP and π angle between QP and PR αExample 4:Consider given tetrahedron. Find the angle between AD and plane BCD.ALet OD be the projection of AD onto BCD.Therefore, AO is perpendicular to plane BCDAngle between AD and BCD Angle ADO cos-1(OD/AD) cos-1(OD/6)66 6B55αODTo find OD consider triangle BCDDSince this is a regular pyramid, O will coincidewith the centroid of triangle BCDOD 2/3 DE 2/3(5 sin 60o) 2.89Therefore, angle ADO cos-1(2.89/6) 61.2o5C251OB2003Three-Dimensional Trigonometry LYWong60oE12C4

4Angle between two planesDefinition:The common line of 2 non-parallel planes π1 and π2 is the line where 2 planes meet.π1AαCCommonlineDPBπ2Consider any point P on the common line CD.PA and PB are drawn at right angles to CD such that PA is in π1 and PB is in π2.Then angle between π1 and π2 is angle APB αExample 5:A triangular pyramid with vertex V and base ABC has VA VB VC 5 cm and AB BC CA 6 cm.(i) Prove that angle between edge VA and base ABC is cos-1(2 3/5).(ii) Calculate the length of perpendicular from B to VA.(iii) Prove that angle between plane faces VAB and VAC is cos-1(7/32)Solution:(i)VAngle between VA and base ABC Angle VAO αcos α OA/5P(ii)γαConsider triangle ABCAE 6 sin 60o 3 3O is the centroid of triangle ABCOA 2/3 AE 2 3Therefore, cos α 2 3/5α cos-1(2 3/5)5ACO6EBV5P5βA3Q2003Three-Dimensional Trigonometry LYWong3B5

Consider triangle VABLet perpendicular from B to VA be BP.cosβ AQ/AV 3/5sin β 4/5BP 6 sin β 24/5(iii) Angle between VAB and VAC is angle BPC γBy symmetry, CP BP 24/5Considering triangle BPC, by cosine rule,222cos γ BP CP BC2 BP CP22 24 24 6 2 5 5 72322 24 5 γ cos 1 7 32 Example 6:Consider the regular pyramid with square base ABCD, (dim. 5 5 cm) and vertex P, length ofeach sloping edge 7 cm. Find,(i) angle between any sloping plane to the base.P(ii) angle between any 2 sloping faces.Solution:7Angle between any sloping face and base angle between BCP and ABCD Angle PEO αConsider triangle BEP,PE2 PB2 - BE2 72 - (5/2)2PE (171/4)OE ½ AB 5/2-1α cos (OE/PE) 67.5o(ii) Angle between 2 sloping faces angle between PCD and PDA Angle AFC βWe need to find FC, FA, AC.7AFBβαODE5CPTo find FC, consider triangle PDCcos ( PDC) DG/DP 5/14sin ( PDC) (171)/14FC 5 sin PDC 4.7By symmetry, FA FCConsidering triangle ADC,AC2 52 52 50AC 502003Three-Dimensional Trigonometry LYWongFD5/27G5/2C6

Considering triangle AFC, by cosine rulecos β ( 4.7) 2 ( 4.7) 2 502( 4.7) 2β 98.4 oExample 7:A square PQRS of side 10 cm lies on a horizontal plane. A second square PQXY of side 10cm also lies in a plane inclined at an angle of 40o to the square PQRS. Find the angles madewith the square PQRS by (i) PX and (ii) the plane PXS.YXPBα40oQβSARSolution:Angle between PX and PQRS angle XPAConsidering square PQXY,PX2 102 102PX 10 2Consider triangle QAX, AX 10 sin 40osin α AX/PX 10 sin 40o/10 2 α 27.03o(ii) Angle between PQRS and PXS angle ABX βtan β AX/AB 10 sin 40o/10β 32.73oExample 8:Each cross-section of a prism is a sector of a circle, of radius 4 cm, with angle at the centreequal to 45o. Two cross-sections are OAB and PDC where A, B, C, D lie on the curved surfaceof the prism and the vertical line OP is the intersection of the vertical plane face OADP andOBCP, as shown in the figure. The lines AD and BC are vertical. The cross-sections OAB andPDC are horizontal and 5 cm apart. Giving each answer correct to 3 significant figures, find(i) the area of the curved surface ABCD.(ii) the angle between AC and the plane OAB.(iii) the angle between the planes OAC and OADP.(iv) the area of the triangle OAC.2003Three-Dimensional Trigonometry LYWong7

CSolution:4EG // DA(i)Area of curved surface 5 length of arc AB 5 4 (π/4) 15.7 cm2(ii) Angle between AC and plane OAB angle BAC αtan α BC/AB 5/ABConsider triangle OAB,AB 2 BF 2(4 sin 22.5o) 3.06 cmtan α 5/3.06 α 58.5oBGPD5βBαo45OEA4O22.5oFA(iii) Angle between OAC and OADP angle CEG βtan β CG/EG CG/5Considering triangle PCG,CG 4 sin 45otan β 4 sin 45o/5β 29.5o(iv) Area of triangle OAC ½ OA ECConsider triangle CEG, EC 5/cos β 5.74 cmArea of triangle OAC ½ 4 5.74 11.48 cm2Example 9:The sides of the square ABCD are each of length a. The rectangle BKLC lies in a planeperpendicular to the plane ABCD and BK CL 2a. Find each of the following angles, givingyour answers to the nearest tenth of a degree.(a) the angle between the line AL and the plane AKB(b) the angle AKC(c) the angle between the planes ACK and ABCD(d) the angle between the skew lines AD and KC.Solution:(a) AK2 AB2 BK2 a2 (2a)2 5a2AK a 5Angle between line AL and plane AKB Angle between line AL and line AK2003Three-Dimensional Trigonometry LYWong8

angle LAK tan-1(LK/AK) tan-1(a/a 5) 24.1oLLKCKDαACD2a2aaBAaB(b) AC2 AB2 BC2 a2 a2 2a2AC a 2KC2 BC2 BK2 a2 (2a)2 5a2KC a 5cos AKC AK 2 KC 2 AC 22 AK KC 5a 2 5a 2 2a 22 5a 2 0 .8 AKC 36.9 o(c) Let α be the angle between plane ACK and ABCDα tan-1(BK/½ BD) tan-1(2a/½a 2) 70.5o (since BD2 a2 a2 2a2)(d) Angle between AD and KC Angle between BC and KC tan-1(2a/a) 63.43oExample 10:A pyramid has a rectangular base ABCD and vertex V; VN is perpendicular to the base ABCDwhile VP, VQ, VR and VS respectively are perpendicular to the sides AB, BC, CD and DA ofbase. PR and QS meet at right angles at N. If AB 18 units, BC 16 units, VA 22 units, VB 14 units, VC 6 units(i) Calculate the lengths of BP and BQ(ii) Prove that VD 18 units and VN 26 units.(iii) Calculate the angle between the plane VDA and base ABCD, correct to the nearest 0.1oSolution:2003Three-Dimensional Trigonometry LYWong9

VV(i)22A1418 - xPxDBRS222 - (18 - x)2 142 - x2484 - 324 36x -x2 142 - x2x 1 BPCNAQPBV14142 - y2 62 - (16 - y)2 36 - 256 32y - y2y 13 BQB(ii)6yQ 16 - yCVDR AP 17z2 - 172 62 - 12z 18VD 18 unitszD617R1CDDS CQ 3DN2 32 172 298317SNVVN2 182 - 298VN 26 units(iii)18 298DVAngle between plane VDAand base ABCD θ tan-1( 26/17) 16.7o 26θS2003Three-Dimensional Trigonometry LYWongN17N10

Example 11:The horizontal base of a pyramid is a rectangle ABCD, where AB 6 cm and BC 8 cm. Thevertex V of the pyramid is such that VA VB VC VD 12 cm. Giving each answer to thenearest 0.1o, find(i) the angle between the skew lines VA and BC(ii) the inclination of the plane VAB to the horizontal(iii) the angle between the two planes VAD and VBCFind the length of the perpendicular from D to the line VC, correct to 3 significant figures.V12FCBNOAM6D8Solution:(i) Angle between VA and BC Angle between VA and AD VADLet M be the mid-point of AD AMV 90o , AM 4 cm VAD VAM cos-1(4/12) 70.5o(ii) Let O be the centre of the base and N be the mid-point of AB VNA VON 90oVN (VA2 - AN2) (122 - 32) 135Angle of inclination of VAB to horizontal VNO cos-1(4/ 135) 69.6o(iii) Angle between VAD and VBC 2 MVOVM (VA2 - AM2) (122 - 42) 1282 MVO 2 sin-1(OM/VM) 2 sin-1(3/ 128) 30.8oArea of VCD Area of VAB ½ AB VN ½ 6 135 3 135Let F be foot of perpendicular from D to VC½ DF VC Area of VCD½ DF 12 3 135DF 5.86 cm2003Three-Dimensional Trigonometry LYWong11

Example 12:The diagram shows a cube with all edges of unit length. P is a point on the edge FB such thatPF a. Find, in terms of a, the perpendicular distance from E to the line HP. If θ is the obtuseangle between the planes EHP and GHP, show that cos θ -1/(1 a2)If a 1/3, find to the nearest 0.1o,(a) the angle between the line HP and the plane ADHE,(b) the angle between the planes EHP and EHC.HGXSolution:FaEEP (1 a2)HF 2222HP (HF a ) (2 a )Consider area of HEP:½ HP EX ½ EH EP (2 a2) EX (1 a2)DPACBPerpendicular distance from E to HP EX (1 a2)/ (2 a2)EX 2 GX 2 EG 22 EX GX 1 a2 22 2 a2 (Q EX GX) 1 a2 2 2 a2 cos θ HG1 a2 2 a2 2 a21 a2 11 aEM2FaPDC2 a2A(a) Angle between HP and ADHE PHMHM (HE2 EM2) [1 (1/3 )2] 10/3 PHM tan-1(PM/HM) tan-1[1/( 10/3)] 43.5o(b) Angle between EHP and EHC Angle between EP and HC Angle between EP and EB γEP 2 EB 2 PB 2cos γ 2 EP EB 10 2 2 9 3 2HGFED21029BAPCBγ 26.6 o2003Three-Dimensional Trigonometry LYWong12

THREE DIMENSIONAL TRIGONOMETRY Lecturer: Mrs Wong Lai Yong 1 Sine rule and cosine rule A Sine rule : sinC c sinB b sinA a c b B a C Cosine rule : a2 b2 c2 - 2bc cos A b2 a2 c2 - 2ac cos B c2 a2 b2 - 2ab cos C OR : 2ab a b c cosC 2ac a c b B 2bc b c a cosA 2 2 2 2 2 2 2 2 2 cos Example 1: N2002/1/9(i) .