3D Pythagoras and Trigonometry

Imagine a light shining straight down from above. Which line in the base is the shadow of AG?

1. To find the angle between a line and a plane, drop a perpendicular from the top end of the line down to the plane.
2. G is vertically above C, so the line straight down from G meets the base at C. A is already in the base.
3. So the shadow of AG on the base is AC, and the angle is \angle GAC in the right-angled triangle GAC.
4. The perpendicular from G meets the base at C. The extracted triangle shows the required angle \theta=\angle GAC.

5. BD is also a diagonal of the base, but it does not pass through A or C, so it is not the shadow of AG.

AC lies in the base. Which right-angled triangle in the base has AC as its hypotenuse?

1. AC is a diagonal of the base ABCD, so only the base is needed. The height AE is not used.
2. Triangle ABC is right-angled at B, so use Pythagoras' theorem: AC^2=AB^2+BC^2.
3. AC^2=6^2+8^2=36+64=100
4. AC=\sqrt{100}=10 cm

Find the base diagonal AC first, then use the right-angled triangle ACG.

1. First the base diagonal: triangle ABC is right-angled at B, so AC^2=6^2+8^2=100.
2. CG is a vertical edge, so triangle ACG is right-angled at C, with CG=AE=7 cm.
3. By Pythagoras' theorem, AG^2=AC^2+CG^2=100+49=149.
4. AG=\sqrt{149}=12.206\ldots, so AG=12.2 cm to 1 decimal place.

The vertex of the angle is where the two lines meet.

1. The angle between two lines is at the point where they meet. AG and CG meet at G.
2. So G is the middle letter: the angle is \angle AGC.
3. \angle CAG is at the other end of AG. It is the angle between AG and the base diagonal AC, not CG.

Find AC^2 in the base first, then use triangle ACG. Keep the answer as a surd.

1. Triangle ABC is right-angled at B: AC^2=5^2+5^2=50.
2. CG is vertical, so triangle ACG is right-angled at C: AG^2=AC^2+CG^2=50+25=75.
3. So AG=\sqrt{75}. To simplify, write 75 using its largest square factor: 75=25\times3.
4. \sqrt{75}=\sqrt{25}\times\sqrt3=5\sqrt3, so AG=5\sqrt3 cm.

The angle is \angle GAC, at A in the right-angled triangle GAC. Find AC first.

1. G is vertically above C, so the angle between AG and the base is \angle GAC. Triangle GAC is right-angled at C.
2. Base diagonal (triangle ABC, right angle at B): AC=\sqrt{9^2+4^2}=\sqrt{97} cm. Keep this exact.
3. In triangle GAC, CG=AE=6 cm is opposite the angle and AC is adjacent to it, so use tan.
4. In the extracted triangle, CG=6 cm is opposite \theta and the base projection AC=\sqrt{97} cm is adjacent.

5. \tan(\angle GAC)=\dfrac{6}{\sqrt{97}}, so \angle GAC=\tan^{-1}\left(\dfrac{6}{\sqrt{97}}\right)=31.35\ldots^\circ
6. The angle is 31.4^\circ to 1 decimal place.

The angle is at G, where AG and CG meet. Find AC first.

1. AG and CG meet at G, so the angle is \angle AGC. Triangle AGC is right-angled at C.
2. Base diagonal (triangle ABC, right angle at B): AC=\sqrt{5^2+12^2}=\sqrt{169}=13 cm.
3. Looking from G: AC=13 cm is opposite the angle and CG=AE=8 cm is adjacent, so use tan.
4. At G, the base diagonal AC=13 cm is opposite \theta and the vertical edge CG=8 cm is adjacent.

5. \tan(\angle AGC)=\dfrac{13}{8}, so \angle AGC=\tan^{-1}\left(\dfrac{13}{8}\right)=58.39\ldots^\circ
6. The angle is 58.4^\circ to 1 decimal place.

Use Pythagoras' theorem twice (the base diagonal, then triangle ACG), then rearrange for the height.

1. Base diagonal: AC^2=10^2+6^2=136.
2. Triangle ACG is right-angled at C, with hypotenuse AG=17 cm and CG=AE, so AG^2=AC^2+AE^2.
3. 17^2=136+AE^2, so AE^2=289-136=153.
4. AE=\sqrt{153}=12.36\ldots, so the height is 12.4 cm to 1 decimal place.

Use trigonometry in the right-angled triangle ABC to find AC first.

1. Triangle ABC is right-angled at B. AB is adjacent to the 34^\circ angle and AC is the hypotenuse, so use cos.
2. \cos34^\circ=\dfrac{9}{AC}, so AC=\dfrac{9}{\cos34^\circ}=10.855\ldots cm
3. G is vertically above C, so triangle ACG is right-angled at C, with CG=AE=4 cm.
4. AG^2=AC^2+CG^2=10.855\ldots^2+4^2=133.85\ldots
5. AG=\sqrt{133.85\ldots}=11.569\ldots, so AG=11.6 cm to 1 decimal place.

The angle with the base is \angle GAC. Find AC, then use tan to find CG.

1. G is vertically above C, so the 27^\circ angle is \angle GAC, in triangle GAC with a right angle at C.
2. Base diagonal: AC=\sqrt{12^2+7^2}=\sqrt{193} cm.
3. CG is opposite the 27^\circ angle and AC is adjacent, so \tan27^\circ=\dfrac{CG}{\sqrt{193}}.
4. CG=\sqrt{193}\times\tan27^\circ=7.078\ldots cm
5. The height AE is equal to CG, so AE=7.1 cm to 1 decimal place.

The longest rod reaches from one corner to the opposite corner. Put all three lengths in the same units first.

1. Use the same units: 0.48 m =48 cm.
2. The longest straight rod fits along a diagonal from a bottom corner to the opposite top corner.
3. Diagonal of the base: d^2=48^2+36^2=2304+1296=3600, so d=60 cm.
4. The rod, the base diagonal and the 20 cm height make a right-angled triangle: L^2=60^2+20^2=4000.
5. L=\sqrt{4000}=63.24\ldots, so the greatest length is 63.2 cm to 1 decimal place.

Use the volume to find the height of the cuboid first.

1. Volume = length \times width \times height, so the height is AE=864\div(12\times9)=864\div108=8 cm.
2. Base diagonal: AC=\sqrt{12^2+9^2}=\sqrt{225}=15 cm.
3. The angle between AG and the base is \angle GAC, in triangle GAC with a right angle at C. CG=8 cm is opposite and AC=15 cm is adjacent.
4. \tan(\angle GAC)=\dfrac{8}{15}, so \angle GAC=28.07\ldots^\circ
5. The angle is 28.1^\circ to 1 decimal place.

Use the perpendicular from the apex to the centre of the base.

1. O is the centre of the square base, so to get from O to the corner A you go half a side (5 cm) one way and half a side (5 cm) the other way.
2. So OA^2=5^2+5^2=50.
3. PO is vertical, so triangle POA is right-angled at O, with hypotenuse PA: PA^2=PO^2+OA^2=144+50=194.
4. After finding OA^2=50, use the vertical height PO and this horizontal distance to find the hypotenuse PA.

5. PA=\sqrt{194}=13.92\ldots, so PA=13.9 cm to 1 decimal place.

Use the perpendicular from the apex to the centre of the base.

1. P is vertically above O, so the angle between PA and the base is \angle PAO, in triangle PAO with a right angle at O.
2. From A to the centre O is half of AB one way and half of BC the other: AO=\sqrt{6^2+4^2}=\sqrt{52} cm.
3. The projection of PA onto the base is AO, so the required angle is at A in this vertical triangle.

4. PO=9 cm is opposite the angle and AO is adjacent, so \tan(\angle PAO)=\dfrac{9}{\sqrt{52}}.
5. \angle PAO=51.29\ldots^\circ, so the angle is 51.3^\circ to 1 decimal place.

Use the perpendicular from the apex to the centre of the base.

1. From the centre O to the corner A is 5 cm one way and 5 cm the other, so OA^2=5^2+5^2=50.
2. Triangle POA is right-angled at O, and the sloping edge PA is the hypotenuse, so PO^2=PA^2-OA^2.
3. PO^2=13^2-50=169-50=119
4. PO=\sqrt{119}=10.90\ldots, so the height is 10.9 cm to 1 decimal place.

Which corner of the face ABFE is directly in front of G? The angle is between AG and the line from A to that corner.

1. The edge FG is perpendicular to the front face ABFE, so the perpendicular from G to that face meets it at F.
2. So the angle between AG and the face ABFE is \angle GAF, in triangle GAF with a right angle at F.
3. AF is a diagonal of the square face ABFE: AF=\sqrt{6^2+6^2}=\sqrt{72}=6\sqrt2 cm. Also FG=6 cm.
4. The extracted triangle shows AF, the projection of AG onto the front face. The required angle is \theta=\angle GAF.

5. FG is opposite the angle and AF is adjacent: \tan(\angle GAF)=\dfrac{6}{6\sqrt2}=\dfrac{1}{\sqrt2}.
6. \angle GAF=35.26\ldots^\circ, so the angle is 35.3^\circ to 1 decimal place.

G is vertically above C, so the angle with the base is \angle GMC. Find MC first.

1. G is vertically above C, so the angle between MG and the base is \angle GMC, in triangle GMC with a right angle at C.
2. M is the midpoint of AB, so MB=4 cm. Triangle MBC is right-angled at B: MC=\sqrt{4^2+6^2}=\sqrt{52} cm.
3. The base projection starts at M, so the required triangle is MCG with horizontal side MC=\sqrt{52} cm.

4. CG=AE=4 cm is opposite the angle and MC is adjacent: \tan(\angle GMC)=\dfrac{4}{\sqrt{52}}.
5. \angle GMC=29.01\ldots^\circ, so the angle is 29.0^\circ to 1 decimal place.

Each side of triangle ACF is a diagonal of a face. Triangle ACF has no right angle, so use the cosine rule.

1. Each side of triangle ACF is a diagonal of one face of the cuboid, so find each one with Pythagoras' theorem. F is vertically above B, so BF=AE=5 cm.
2. Base ABCD: AC^2=10^2+6^2=136.
3. Side face BCGF: CF^2=BC^2+BF^2=6^2+5^2=61.
4. Front face ABFE: AF^2=AB^2+BF^2=10^2+5^2=125.
5. These three face diagonals form triangle ACF. The angle \theta is opposite AF, so AF^2 is the term subtracted in the cosine rule.

6. Triangle ACF has no right angle, so use the cosine rule for the angle at C: \cos C=\dfrac{AC^2+CF^2-AF^2}{2\times AC\times CF}
7. \cos(\angle ACF)=\dfrac{136+61-125}{2\times\sqrt{136}\times\sqrt{61}}=\dfrac{72}{182.16\ldots}=0.3952\ldots
8. \angle ACF=\cos^{-1}(0.3952\ldots)=66.71\ldots^\circ, so the angle is 66.7^\circ to 1 decimal place.

Use triangle POM to find the height PO first.

1. M is the midpoint of BC and O is the centre, so OM is half of AB: OM=7 cm.
2. PO is vertical, so triangle POM is right-angled at O: PO^2=PM^2-OM^2=144-49=95.
3. The midpoint distance OM=7 cm and sloping face height PM=12 cm give the vertical height PO.

4. From O to the corner A is 7 cm one way and 7 cm the other, so OA^2=7^2+7^2=98.
5. The angle between PA and the base is \angle PAO, with a right angle at O: \tan(\angle PAO)=\dfrac{PO}{OA}=\dfrac{\sqrt{95}}{\sqrt{98}}.
6. For the angle of PA to the base, use the corner distance OA=\sqrt{98} cm, with the same height PO=\sqrt{95} cm.

7. \angle PAO=44.55\ldots^\circ, so the angle is 44.6^\circ to 1 decimal place.

Find the height PO from triangle POA, then use triangle PMO.

1. From O to the corner A is half of AB one way and half of BC the other: OA^2=6^2+5^2=61.
2. In the base, OA runs to a corner and OM runs to the midpoint of a side. They are different lengths.

3. PO is vertical, so in triangle POA (right angle at O): PO^2=PA^2-OA^2=225-61=164.
4. Use the corner distance OA with the sloping edge PA to find the vertical height PO.

5. M is the midpoint of AB and O is the centre, so OM is half of BC: OM=5 cm.
6. Triangle PMO is right-angled at O. PO is opposite \angle PMO and OM is adjacent: \tan(\angle PMO)=\dfrac{\sqrt{164}}{5}.
7. Now use OM=5 cm, rather than the corner distance OA. In triangle PMO, the height PO is opposite \theta=\angle PMO.

8. \angle PMO=68.67\ldots^\circ, so the angle is 68.7^\circ to 1 decimal place.