Angles in 3D shapes Problem set A One question at a time

ABCDEFGH is a cuboid with AB = 7 cm, BC = 4 cm and AE = 5 cm. Find angle FAC.

• \(AF^2 = 7² + 5² = 74\), \(AC^2 = 7² + 4² = 65\), \(FC^2 = 4² + 5² = 41\)
• \(\cos FAC = \dfrac{74 + 65 - 41}{2\sqrt{74}\sqrt{65}} = 0.7065\)
• \(\angle FAC = 45.0^\circ\)

ABCDEFGH is a cuboid with AB = 8 cm, BC = 6 cm and AE = 3 cm. The diagonals AG and EC cross at O. Find the acute angle between AG and EC.

• \(AG^2 = 8^2 + 6^2 + 3^2 = 109\), so \(AO = OC = \tfrac12\sqrt{109}\) (the diagonals cross at their midpoints)
• \(AE^2 = 3² = 9\)
• \(\cos AOE = \dfrac{27.25 + 27.25 - 9}{2 \times 27.25} = 0.8349\)
• \(\angle AOE = 33.40^\circ\), which is acute

ABCDE is a pyramid. The base ABCD is a square of side 6 cm. Each sloping edge is 10 cm.

1. (a) Find angle EBC.

   • \(\cos \angle EBC = \dfrac{3}{10}\) (BM is half of BC, in the isosceles triangle EBC)
   • \(\angle EBC = 72.5^\circ\)

2. (b) Find angle EBD.

   • \(BD = \sqrt{6^2 + 6^2}\), so \(BN = 4.243\)
   • \(\cos \angle EBD = \dfrac{4.243}{10}\), \(\angle EBD = 64.9^\circ\)

3. (c) Find the height of the pyramid.

   • \(h^2 = 10^2 - 4.243^2 = 82.00\)
   • \(h = 9.06\) cm

ABCDEFGH is a cuboid with AB = 9 cm, BC = 5 cm and AE = 3 cm. Find the angle between the diagonal AG and the base ABCD.

• \(AC^2 = 9^2 + 5^2 = 106\)
• \(\tan \theta = \dfrac{CG}{AC} = \dfrac{3}{\sqrt{106}}\)
• \(\theta = 16.2^\circ\)

ABCDEFGH is a cuboid with AB = 8 cm, BC = 3 cm and AE = 5 cm. Find the angle between the diagonal AG and the face BCGF.

• \(AB\) is perpendicular to the face BCGF, so the shadow of AG on the face is BG. The angle is at G.
• \(BG^2 = 3^2 + 5^2 = 34\)
• \(\tan \theta = \dfrac{AB}{BG} = \dfrac{8}{\sqrt{34}}\), \(\theta = 53.9^\circ\)

A storage crate is a cuboid 120 cm long, 80 cm wide and 60 cm high. A straight rod rests with one end at corner A on the floor and the other at the opposite top corner G.

1. (a) Find the length of the longest rod that fits, which is the diagonal AG.

   • \(AG^2 = 120^2 + 80^2 + 60^2 = 24400\)
   • \(AG = 156\) cm

2. (b) Find the angle between this rod and the base.

   • \(AC^2 = 120^2 + 80^2 = 20800\)
   • \(\tan \theta = \dfrac{60}{\sqrt{20800}}\)
   • \(\theta = 22.6^\circ\)

A tent is a square-based pyramid. The square base has side 4 m and each sloping edge is 3.6 m long.

1. (a) Find the height of the pyramid.

   • \(BN = \tfrac12 \times 4\sqrt{2} = 2.828\)
   • \(h^2 = 3.6^2 - 2.828^2 = 4.960\)
   • \(h = 2.23\) m

2. (b) Find the angle between a sloping edge and the base.

   • \(\cos \theta = \dfrac{BN}{EB} = \dfrac{2.828}{3.6}\)
   • \(\theta = 38.2^\circ\)

A glass ornament is a pyramid ABCD. The base ABC is an equilateral triangle of side 10 cm. The point D is directly above M, the centre of the base, and AD = 14 cm.

1. (a) Find the angle between AD and the base.

   • \(\cos 30^\circ = \dfrac{5}{AM}\), so \(AM = 5.774\)
   • \(\cos \theta = \dfrac{5.774}{14}\), \(\theta = 65.6^\circ\)

2. (b) Find the height DM.

   • \(DM^2 = 14^2 - 5.774^2 = 162.7\)
   • \(DM = 12.8\) cm

A box has a square base ABCD of side 6 cm. The diagonal AG makes an angle of 35° with the base. Find the height of the box, h.

• \(AC^2 = 6^2 + 6^2 = 72\)
• \(\tan 35^\circ = \dfrac{h}{\sqrt{72}}\)
• \(h = \sqrt{72} \tan 35^\circ = 5.94\) cm