THREE DIMENSIONAL TRIGONOMETRY Lecturer: Mrs Wong Lai Yong 1 Sine rule and cosine rule A

Sine rule : Csin

cBsin

bAsin

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 :

ab2cbaCcos

ac2bcaB

bc2acbAcos

222

222

222

−+=

−+=

−+=

cos

Example 1: N2002/1/9(i)

In triangle ABC, angle B = θ, angle C = θ + α, AB = 2 and AC = 1. Show that α−

α=θ

cos2sintan .

Solution:

θ + α B

2 1

θ

A

2)sin(sin α+θ

=θ

2 sin θ = sin θ cos α + cos θ sin α 2 tan θ = tan θ cos α + sin α tan θ (2 - cos α) = sin α

C α−

α=θ

cos2sintan

Example 2: J95/1/4 The 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 is

4

26 − . O

A B

1 √2

30o

Solution: (a) Let angle ABO = θ

C

2003Three-Dimensional Trigonometry_LYWong 1

o

o

o

1352230sin2sin

sin2

30sin1

=θ∴

==θ

θ=

angle AOB = 180o - 135o - 30o = 15o

oo

o

15sin215sin222AB

sin2

15sinAB

=××=

θ=

(b) 2

AC30cos o =

AC = √2 x (√3/2) = √6/2 BC = AC - AB = √6/2 - 2 sin 15o But cos ∠OBC = BC/1 ∴BC = cos 45o = √2/2 Hence, √2/2 = √6/2 - 2 sin 15o

2 sin 15o = 2

26 −

∴sin 15o = 4

26 −

2 Angle between two lines 2.1 Lines in common plane These are either parallel, non-parallel or intersecting

β

α

L1 L2

Angle between L1 and L2 is α or β.

α β A

B C

Angle between AC and AB is β. Angle between CA and AB is α.

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2.2 Lines lying in different planes These are non-parallel and non-intersecting. They are called skew lines.

E F H G

A B α

D C AG 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 two skew 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 = 90o Example 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.

γ

12

C G

E

D 3 H

4 α β

A 12 B Solution: (i) tan α = ¾ α = 36.87o

F (ii) DG2 = 32 + 122 DG = 12.35 tan β = 12.35/4 β = 72.08o (iii) Angle between DG and EF = Angle between DG and HG ( since HG parallel to EF) = γ tan γ = 3/12 γ = 14.04o

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3 Angles between a line and a plane

Q

α P π R

Line QP meets plane π at P. Drop a perpendicular from Q onto π. Call this QR. Then, RP is the projection of QP onto π. Definition: Angle between a line and a plane = Angle between a line and its projection onto the plane Therefore, angle between QP and π = angle between QP and PR = α Example 4: Consider given tetrahedron. Find the angle between AD and plane BCD.

D

O 5

1 2

1

2

O

5

D α

60o

C

5

B

5

6 6 6

A Let OD be the projection of AD onto BCD. Therefore, AO is perpendicular to plane BCD Angle between AD and BCD = Angle ADO = cos-1(OD/AD) =cos-1(OD/6) To find OD consider triangle BCD Since this is a regular pyramid, O will coincide with the centroid of triangle BCD OD = 2/3 DE =2/3(5 sin 60o) = 2.89 Therefore, angle ADO = cos-1(2.89/6) = 61.2o B E C 2003Three-Dimensional Trigonometry_LYWong 4

4 Angle between two planes Definition: The common line of 2 non-parallel planes π1 and π2 is the line where 2 planes meet.

π2

D

B

α

P C

A π1 Common line

Consider 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 = α Example 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:

6

O

α γ

P

A

E

5

V (i) Angle between VA and base ABC = Angle VAO = α cos α = OA/5 Consider triangle ABC

C AE = 6 sin 60o = 3√3 O is the centroid of triangle ABC OA = 2/3 AE = 2√3 Therefore, cos α = 2√3/5 α = cos-1(2√3/5)

B (ii) V

β

5 P 5

A 3 Q 3 B

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Consider triangle VAB Let perpendicular from B to VA be BP. cosβ = AQ/AV = 3/5 sin β = 4/5 BP = 6 sin β = 24/5 (iii) Angle between VAB and VAC is angle BPC = γ By symmetry, CP = BP = 24/5 Considering triangle BPC, by cosine rule,

=γ∴

=

−

+

=

××−+=γ

−327cos

327

5242

6524

524

CPBP2BCCPBPcos

1

2

222

222

Example 6: Consider the regular pyramid with square base ABCD, (dim. 5 × 5 cm) and vertex P, length of each sloping edge 7 cm. Find, (i) angle between any sloping plane to the base. P (ii) angle between any 2 sloping faces.

7

O

A

α E

7

F β

Solution: Angle between any sloping face and base = angle between BCP and ABCD =Angle PEO = α Consider triangle BEP, PE2 = PB2 - BE2 B =72 - (5/2)2 PE = √(171/4) OE = ½ AB = 5/2 α = cos-1(OE/PE) = 67.5o (ii) Angle between 2 sloping faces = angle between PCD and PDA D 5 C = Angle AFC = β We need to find FC, FA, AC.

6

P To find FC, consider triangle PDC cos (∠PDC) = DG/DP = 5/14 F 7 sin (∠PDC) = √(171)/14 FC = 5 sin ∠PDC = 4.7 By symmetry, FA = FC Considering triangle ADC, AC2 = 52 + 52 = 50 AC = √50

D 5/2 G 5/2 C 2003Three-Dimensional Trigonometry_LYWong

Considering triangle AFC, by cosine rule

o

2

22

4.98

)7.4(250)7.4()7.4(cos

=β

−+=β

Example 7: A square PQRS of side 10 cm lies on a horizontal plane. A second square PQXY of side 10 cm also lies in a plane inclined at an angle of 40o to the square PQRS. Find the angles made with the square PQRS by (i) PX and (ii) the plane PXS.

S

B β

P α

A R

Y X Solution: Angle between PX and PQRS = angle XPA Considering square PQXY, PX2 = 102 + 102 PX = 10√2 Consider triangle QAX, AX = 10 sin 40o sin α = AX/PX = 10 sin 40o/10√2 ∴ α = 27.03o (ii) Angle between PQRS and PXS = angle ABX = β tan β = AX/AB = 10 sin 40o/10 β = 32.73o Example 8: Each cross-section of a prism is a sector of a circle, of radiequal to 45o. Two cross-sections are OAB and PDC whereof the prism and the vertical line OP is the intersection of thOBCP, as shown in the figure. The lines AD and BC are vPDC are horizontal and 5 cm apart. Giving each answer c(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_LYWong

40o

Q

us 4 cm, with angle at the centre A, B, C, D lie on the curved surface e vertical plane face OADP and

ertical. The cross-sections OAB and orrect to 3 significant figures, find

7

β

4

α

B

G P

C

45o O E A

Solution: EG // DA (i) Area of curved surface = 5 × length of arc AB D =5 × 4 (π/4) = 15.7 cm2 (ii) Angle between AC and plane OAB = angle BAC = α tan α = BC/AB = 5/AB

5 Consider triangle OAB, AB = 2 BF = 2(4 sin 22.5o) = 3.06 cm tan α = 5/3.06 ⇒ α = 58.5o B

F A

22.5o

4

O

(iii) Angle between OAC and OADP = angle CEG = β tan β = CG/EG = CG/5 Considering triangle PCG, CG = 4 sin 45o tan β = 4 sin 45o/5 β = 29.5o (iv) Area of triangle OAC = ½ × OA × EC Consider triangle CEG, EC = 5/cos β = 5.74 cm Area of triangle OAC = ½ × 4 × 5.74 = 11.48 cm2 Example 9: The sides of the square ABCD are each of length a. The rectangle BKLC lies in a plane perpendicular to the plane ABCD and BK = CL = 2a. Find each of the following angles, giving your 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 = 5a2

AK = a√5 Angle between line AL and plane AKB = Angle between line AL and line AK

2003Three-Dimensional Trigonometry_LYWong 8

=angle LAK

A a

C

L

D

=tan-1(LK/AK) = tan-1(a/a√5) = 24.1o

α

A a

C

L

D

K 2a B

K

2a B

(b) AC2 = AB2 + BC2 = a2 + a2 = 2a2 AC = a√2 KC2 = BC2 + BK2 = a2 + (2a)2 = 5a2 KC = a√5

o

2

222

222

9.36AKC

8.0a52

a2a5a5

KCAK2ACKCAKAKCcos

=∠

=×

−+=

××−+

=∠

(c) Let α be the angle between plane ACK and ABCD α = 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) = 63.43o Example 10: A pyramid has a rectangular base ABCD and vertex V; VN is perpendicular to the base ABCD while VP, VQ, VR and VS respectively are perpendicular to the sides AB, BC, CD and DA of base. 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 0.1o Solution:

2003Three-Dimensional Trigonometry_LYWong 9

A P B

S N Q

V

D R C A 18 - x P x B

22 14

V (i) 222 - (18 - x)2 = 142 - x2 484 - 324 + 36x -x2 = 142 - x2 x = 1 = BP

B y Q 16 - y C

14 6

V 142 - y2 = 62 - (16 - y)2 = 36 - 256 + 32y - y2 y = 13 = BQ (ii)

z 6

V DR = AP = 17 z2 - 172 = 62 - 12 z = 18 VD = 18 units D 17 R 1 C

D 3 S

DS = CQ = 3 DN2 = 32 + 172 = 298

17 N

18

D √298

V N

VN2 = 182 - 298 VN = √26 units (iii)

θ S 17

V √26 N

Angle between plane VDA and base ABCD = θ = tan-1(√26/17)=16.7o 2003Three-Dimensional Trigonometry_LYWong 10

Example 11: The horizontal base of a pyramid is a rectangle ABCD, where AB = 6 cm and BC = 8 cm. The vertex V of the pyramid is such that VA = VB = VC = VD = 12 cm. Giving each answer to the nearest 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 VBC Find the length of the perpendicular from D to the line VC, correct to 3 significant figures. V

12

B

F C

N O 6

A M D 8

Solution: (i) Angle between VA and BC =Angle between VA and AD = ∠ VAD Let 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 = 90o VN = √(VA2 - AN2) = √(122 - 32) = √135 Angle of inclination of VAB to horizontal = ∠VNO = cos-1(4/√135) = 69.6o (iii) Angle between VAD and VBC = 2×√MVO VM = √(VA2 - AM2) = √(122 - 42) = √128 2×√MVO = 2 sin-1(OM/VM) =2 sin-1(3/√128) = 30.8o Area of ∆VCD = Area of ∆VAB = ½ × AB × VN = ½ × 6 × √135 = 3√135 Let F be foot of perpendicular from D to VC ½ × DF× VC = Area of ∆VCD ½ × DF× 12= 3√135 DF = 5.86 cm 2003Three-Dimensional Trigonometry_LYWong 11

Example 12: The diagram shows a cube with all edges of unit length. P is a point on the edge FB such that PF = a. Find, in terms of a, the perpendicular distance from E to the line HP. If θ is the obtuse angle 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.

H G

F E a D P C A B

X

Solution: EP = √(1 + a2) HF = √2 HP = √(HF2 + a2) = √(2 + a2) Consider area of ∆HEP: ½ × HP × EX = ½ × EH × EP √(2 + a2) × EX = √(1 + a2) Perpendicular distance from E to HP = EX = √(1 + a2)/ √(2 + a2)

2

2

2

2

22

2

2

2

2

222

a11

a2a1a2

a2a1

)GXEX(

a2a12

2a2a12

GXEX2EGGXEXcos

+−=

+

++

−−+

=

=

+

+

−

+

+

=

××−+

=θ

Q H G

F E a M D P C A B

(a) Angle between HP and ADHE = ∠ PHM HM = √(HE2 + EM2) = √[1 + (1/3 )2] = √10/3 ∠ PHM = tan-1(PM/HM) = tan-1[1/(√10/3)] = 43.5o H G (b) Angle between EHP and EHC

F E D P C A B

= Angle between EP and HC = Angle between EP and EB = γ

o

2

222

6.26

29

102

322

910

EBEP2PBEBEPcos

=γ

−+

=

××−+

=γ

2003Three-Dimensional Trigonometry_LYWong 12