3 D problem 3 First created by Mr. Francis Hung on 20120728

In the figure, VABCDE is a right pentagonal pyramid with a regular pentagonal base ABCDE with side = 2. V is the vertex of the pyramid and O is the projection of V on the base. VA = VB = VC = VD = VE = 2. Find the angle between (a) VB and the base, (b) the plane VBC and the base, (c) the planes VAB and VBC, (d) CB and the plane VAB. By definition, O is the centre of the regular pentagon. AB = BC = CD = DE = EA = 2 ∠BAE = ∠ABC = ∠BCD = ∠CDE = ∠AED = 108° (∠s sum of polygon) Join OA, OB, OC, OD, OE. OA = OB = OC = OD = OE ∆OAB ≅ ∆OBC ≅ ∆OCD ≅ ∆ODE ≅ ∆OEA (S.S.S.) ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠AOE = 72° (∠s at a pt., corr. ∠s ≅ ∆s) ∠OAB = ∠OBC (Q OA = OB, base, ∠s isos. ∆) =

Last updated: November 3, 2015

D

C

E O

A

G

B

180 o − ∠AOB = 54° (∠s sum of ∆) 2

Let G be the projection of O on AB. Then OG ⊥ AB, ∆OAG ≅ ∆OBG (R.H.S.) AG = GB = 1 (corr. sides ≅ ∆s), OG = GB tan 54° = tan 54°, OA = OB = OC = OD = OE = sec 54° ∆VAB, ∆VBC, ∆VCD, ∆VDE, ∆VAE are congruent equilateral triangles with side = 2. (a) Join VO and OB. cos ∠OBV =

OB sec 54 o = VB 2

∠OBV = 31.7° (correct to 3 sig. fig.) (b)

Let M be the mid point of BC. BM = MC = 1. ∆VBM ≅ ∆VCM (S.S.S.) ∠BMV = ∠CMV = 90° (corr. ∠s ≅ ∆s, adj. ∠s on st. line) By definition, ∠VOM = 90° (O is the projection of V) In ∆VBM, BM2 + VM2 = VB2 (Pythagoras’ theorem) ∴ VM = 3 In ∆OBM, OM = BM tan 54° = tan 54° In ∆VOM, cos ∠VMO =

(c)

OM tan 54 o = VM 3

∠VMO = 37.4° (correct to 3 sig. fig.) Let N be the mid point of VB, VN = NB = 1. Join AN, NC, AC. ∆VAN ≅ ∆BAN (S.S.S.) ∠VNA = ∠BNA = 90° (corr. ∠s ≅ ∆s, adj. ∠s on st. line) ∆VCN ≅ ∆BCN (S.S.S.) ∠VNC = ∠BNC = 90° (corr. ∠s ≅ ∆s, adj. ∠s on st. line) Similar to (b), AN = 3 = NC In ∆ABC, AC = 2×2 sin54° = 4 sin 54° 1 2

In ∆ANC, sin ∠ANC =

1 2

AC

AN

=

2 sin 54 o 3

∠ANC = 138° (correct to 3 sig. fig.) http://www.hkedcity.net/ihouse/fh7878/

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3D problem 3 (d)

Created by Mr. Francis Hung

In ∆VOC, VO2 = VC2 – OC2 (Pythagoras’ theorem) VO = 4 − sec 2 54 o 1 Area of ∆ABC = AB×BC sin 108° = 2 sin 108° 2 Volume of the tetrahedron VABC 1 = ×Area of ∆ABC×VO 3 1 = ×2 sin 108°× 4 − sec 2 54 o …… (1) 3 Let F be the projection of C on the plane VAB. 1 Area of ∆VAB = 2×2 sin 60° = 3 2 Volume of the tetrahedron VABC 1 = ×Area of ∆VAB×CF 3 1 = × 3 ×CF …… (2) 3 1 1 (1) = (2) ⇒ ×2 sin 108°× 4 − sec 2 54 o = × 3 ×CF 3 3 CF =

2 sin 108 o 4 − sec 2 54 o 3

CF sin 108 o 4 − sec 2 54 o = BC 3 ∠CBF = 35.2° (correct to 3 sig. fig.)

In ∆BCF, sin ∠CBF =

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3D problem 3

Created by Mr. Francis Hung

The following figure shows a regular dodecahedron with side = 2 (units). ABCDE and ABQUP are two adjacent faces. Find the angle between these two adjacent faces.

ABCDE and ABQUP are regular pentagons with sides = 2. ∠PAB = 108° (∠s sum of polygon) Produce PA, QB to meet at V. Join VC, VD, VE. Then VABCDE is a right pentagonal pyramid with vertex V and a regular pentagonal base ABCDE. ∠VAB = 72° = ∠VBA (adj. ∠s on st. line) ∆VAB ≅ ∆VBC ≅ ∆VCD ≅ ∆VDE ≅ ∆VEA and VA = VB = VC = VD = VE Let O be the projection of V on the base ABCDE. Then O is the centre of ABCDE. Let M be the mid-point of AB. Then ∠VMB = 90° = ∠VOM = ∠AMO and AM = MB = 1 In ∆AOM, ∠BAE = 108°, ∠OAM = 54°, OM = AM tan ∠BAE = tan 54° In ∆VAM, VM = AM tan ∠VAM = tan 72° OM tan 54 o In ∆VMO, cos ∠VMO = = VM tan 72 o ∠VMO = 63.4° (correct to 3 significant figures) The angle between ABCDE and ABQUP is ∠OMU = 180° – 63.4° = 116.6° (adj. ∠s on st. line)

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