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.)
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Page 1
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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Page 2
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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Page 3