Pythagoras and Trigonometry in 3D

The two planes meet along DC: you need a line in each plane that is perpendicular to DC.

1. The plane EFCD and the base meet along the line DC
2. In the base, CB is perpendicular to DC
3. In the plane EFCD, CF is perpendicular to DC (it lies in the face BCGF)
4. So the angle between the planes is the angle between CB and CF
5. Angle FDB is the angle between the line DF and the base; angle FCG is measured from the vertical; angle CFB is the third angle of triangle FBC
6. Answer: angle FCB

The longest rod joins a bottom corner to the opposite top corner, so use Pythagoras' theorem in 3D.

1. The longest rod lies along a diagonal joining opposite corners
2. d^2=4^2+4^2+7^2
3. d^2=16+16+49=81
4. d=\sqrt{81}
5. Answer: 9 cm

AX is half of the diagonal AC of the base, not half of a side.

1. Diagonal of the base: AC^2=12^2+12^2=288
2. AX=\frac12AC, so AX^2=\frac14\times288=72
3. Triangle VXA has a right angle at X
4. VA^2=AX^2+VX^2=72+49=121
5. Answer: VA=11 cm

If each edge is x cm, 3D Pythagoras gives the diagonal squared as x^2+x^2+x^2.

1. Let each edge be x cm
2. 3D Pythagoras: \;x^2+x^2+x^2=12^2
3. 3x^2=144, so x^2=48
4. x=\sqrt{48}=\sqrt{16}\times\sqrt3
5. x=4\sqrt3 cm

The radius, the perpendicular height and the slant height make a right-angled triangle, and the radius is half the diameter.

1. Radius =18\div2=9 cm
2. In the right-angled triangle, the radius is adjacent to the angle and the slant height is the hypotenuse
3. \cos\theta=\dfrac{9}{15}
4. \theta=\cos^{-1}(0.6)=53.13\ldots^\circ
5. Answer: 53.1^\circ

The two faces meet along AB, and BC and BF are both perpendicular to AB.

1. The sloping face and the base meet along AB
2. BC (in the base) and BF (in the sloping face) are both perpendicular to AB
3. So the angle is angle FBC, in triangle FBC with a right angle at C
4. \tan FBC=\dfrac{45}{240}
5. Angle FBC=\tan^{-1}(0.1875)=10.61\ldots^\circ (the 90 cm is not needed)
6. Answer: 10.6^\circ

The two planes meet along AD, so the 22^\circ angle is at A, between AB and AP.

1. The plane ADQP and the base meet along AD
2. AB and AP are both perpendicular to AD, so angle PAB=22^\circ
3. Triangle ABP has a right angle at B
4. \tan22^\circ=\dfrac{BP}{15}
5. BP=15\tan22^\circ=6.060\ldots (the 8 cm is not needed)
6. Answer: 6.06 cm

M is directly above the midpoint N of DC: find AN first, then use the right-angled triangle ANM.

1. Let N be the midpoint of DC, directly below M, so DN=5 cm
2. AN^2=AD^2+DN^2=6^2+5^2=61
3. AM^2=AN^2+NM^2=61+8^2=125
4. AM=\sqrt{125}=\sqrt{25}\times\sqrt5
5. Answer: 5\sqrt5 cm

The face and the base meet along BC, so the angle is at M in the right-angled triangle VXM.

1. The face VBC and the base meet along BC
2. XM and VM are both perpendicular to BC, so the angle is angle VMX
3. XM is half of AB (not half of BC): XM=8 cm
4. \tan VMX=\dfrac{9}{8}
5. Angle VMX=\tan^{-1}(1.125)=48.36\ldots^\circ
6. Answer: 48.4^\circ

G is directly above C, so the angle you want is in the right-angled triangle ACG, at A.

1. G is vertically above C, so the angle is angle GAC
2. Base diagonal: \;AC^2=9^2+4^2=97, so AC=\sqrt{97}
3. Triangle ACG has a right angle at C: opposite =CG=6, adjacent =AC=\sqrt{97}
4. \tan GAC=\dfrac{6}{\sqrt{97}}
5. Angle GAC=\tan^{-1}\!\left(\dfrac{6}{\sqrt{97}}\right)=31.35\ldots
6. 31.4^\circ

The angle is at A in the right-angled triangle VXA, and AX is half of the diagonal AC.

1. Diagonal of the base: AC^2=24^2+10^2=676, so AC=26 cm
2. AX=\frac12AC=13 cm
3. V is above X, so the angle is angle VAX, with a right angle at X
4. \cos VAX=\dfrac{13}{20}
5. Angle VAX=\cos^{-1}(0.65)=49.45\ldots^\circ
6. Answer: 49.5^\circ

Use the right-angled triangle BCT to find the height CT, then you need the diagonal AC of the car park.

1. Triangle BCT has a right angle at C: \;CT=20\tan35^\circ=14.00\ldots m
2. Diagonal of the car park: AC^2=48^2+20^2=2704, so AC=52 m
3. Triangle ACT has a right angle at C: \;\tan TAC=\dfrac{14.00\ldots}{52}
4. Angle TAC=15.07\ldots^\circ
5. Answer: 15.1^\circ

Write an equation in r and l and use it to write l in terms of r.

1. Base area =\pi r^2, so \pi rl=3\pi r^2
2. Divide by \pi r: \;l=3r
3. In the right-angled triangle, the radius is adjacent to the angle and the slant height is the hypotenuse
4. \cos\theta=\dfrac{r}{l}=\dfrac{r}{3r}=\dfrac13
5. \theta=\cos^{-1}\left(\dfrac13\right)=70.52\ldots^\circ
6. Answer: 70.5^\circ

O is the midpoint of both diagonals, so OA=OB=\frac12AG; then use the cosine rule in triangle AOB.

1. AG^2=8^2+6^2+4^2=116
2. OA=OB=\frac12AG, so OA^2=OB^2=\frac14\times116=29
3. Cosine rule in triangle AOB: \;\cos AOB=\dfrac{OA^2+OB^2-AB^2}{2\times OA\times OB}
4. \cos AOB=\dfrac{29+29-64}{2\times29}=-\dfrac{6}{58}
5. Angle AOB=\cos^{-1}\left(-\dfrac{6}{58}\right)=95.93\ldots^\circ
6. Answer: 95.9^\circ

Let N be the point on BC closest to A: the angle you want is angle DNA, and the area of triangle ABC gives you AN.

1. BC^2=8^2+6^2=100, so BC=10 cm
2. Let N be the point on BC with AN perpendicular to BC
3. Area of ABC=\frac12\times8\times6=24, and also =\frac12\times10\times AN, so AN=4.8 cm
4. D is above A, so DN is also perpendicular to BC: the angle is angle DNA
5. \tan DNA=\dfrac{5}{4.8}
6. Angle DNA=46.16\ldots^\circ (N is not the midpoint of BC)
7. Answer: 46.2^\circ

In part (b), AB is perpendicular to the face BCGF, so the angle you want is in triangle ABG.

Part (a)

1. AC^2=20^2+15^2=625, so AC=25 cm
2. G is vertically above C, so angle GAC=29^\circ and angle ACG=90^\circ
3. \tan29^\circ=\dfrac{CG}{25}
4. CG=25\tan29^\circ=13.85\ldots
5. Answer: 13.9 cm

Part (b)

1. AB is perpendicular to the face BCGF, so B is the point of that face closest to A
2. AG meets the face at G, so the angle is angle AGB, with a right angle at B
3. AG=\dfrac{25}{\cos29^\circ}=28.58\ldots cm
4. \sin AGB=\dfrac{AB}{AG}=\dfrac{20}{28.58\ldots}
5. Angle AGB=44.40\ldots^\circ
6. Answer: 44.4^\circ

Use the right-angled triangles ACT and BCT for the distances along the ground, then the cosine rule in triangle ACB. Answer part (a) before attempting part (b).

Part (a)

1. Triangle ACT has a right angle at C
2. \tan18^\circ=\dfrac{45}{AC}
3. AC=\dfrac{45}{\tan18^\circ}=138.49\ldots
4. Answer: 138 m

Part (b)

1. BC=\dfrac{45}{\tan27^\circ}=88.31\ldots m
2. Triangle ACB is horizontal but not right-angled, so use the cosine rule
3. AB^2=AC^2+BC^2-2\times AC\times BC\times\cos115^\circ
4. AB^2=19181.0\ldots+7799.9\ldots+10338.4\ldots=37319.3\ldots
5. AB=193.18\ldots
6. Answer: 193 m

Use the ratio to place P, find AP, PC and AC by Pythagoras, then use the cosine rule in triangle APC. Answer part (a) before attempting part (b).

Part (a)

1. FG=BC=9 cm, so FP=6 cm and PG=3 cm
2. P is directly above the point N on BC with BN=6 cm
3. AN^2=12^2+6^2=180
4. AP^2=AN^2+10^2=280
5. AP=\sqrt{280}=\sqrt4\times\sqrt{70}
6. Answer: 2\sqrt{70} cm

Part (b)

1. PC^2=PG^2+GC^2=3^2+10^2=109
2. AC^2=12^2+9^2=225, so AC=15 cm
3. Cosine rule: \;\cos APC=\dfrac{AP^2+PC^2-AC^2}{2\times AP\times PC}
4. \cos APC=\dfrac{280+109-225}{2\times\sqrt{280}\times\sqrt{109}}=0.4693\ldots
5. Angle APC=62.00\ldots^\circ
6. Answer: 62.0^\circ

F is directly above the midpoint of BC; for part (b) you also need FC and AC for the cosine rule. Answer part (a) before attempting part (b).

Part (a)

1. Let N be the midpoint of BC: F is 1.5 m directly above N
2. AN^2=AB^2+BN^2=6^2+2^2=40
3. AF^2=AN^2+NF^2=40+1.5^2=42.25
4. AF=\sqrt{42.25}
5. Answer: 6.5 m

Part (b)

1. FC^2=NC^2+NF^2=2^2+1.5^2=6.25, so FC=2.5 m
2. AC^2=6^2+4^2=52
3. Cosine rule: \;\cos AFC=\dfrac{AF^2+FC^2-AC^2}{2\times AF\times FC}
4. \cos AFC=\dfrac{42.25+6.25-52}{2\times6.5\times2.5}=-\dfrac{3.5}{32.5}
5. The cosine is negative, so the angle is obtuse: angle AFC=96.18\ldots^\circ
6. Answer: 96.2^\circ

Let N be the point on the base vertically below M: it is the midpoint of XC, and MN is half of VX.

1. Let N be the point on the base directly below M
2. M is halfway up VC, so N is the midpoint of XC and MN=\dfrac12\times14=7 cm
3. Diagonal AC=\sqrt{10^2+10^2}=\sqrt{200}, so XB=XC=\dfrac12\sqrt{200}=\sqrt{50}
4. XN=\dfrac12XC=\dfrac12\sqrt{50}, so XN^2=12.5
5. The diagonals of a square are perpendicular, so angle BXN=90^\circ
6. BN^2=XB^2+XN^2=50+12.5=62.5, so BN=7.905\ldots
7. Angle between BM and the base is angle MBN: \;\tan MBN=\dfrac{7}{7.905\ldots}
8. Angle MBN=\tan^{-1}(0.8854\ldots)=41.52\ldots
9. 41.5^\circ