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

ABCDEFGH is a cuboid with AB = 9 cm, BC = 5 cm and AE = 6 cm. Find angle BDE.

• \(DB^2 = 9² + 5² = 106\), \(DE^2 = 5² + 6² = 61\), \(BE^2 = 9² + 6² = 117\)
• \(\cos BDE = \dfrac{106 + 61 - 117}{2\sqrt{106}\sqrt{61}} = 0.3109\)
• \(\angle BDE = 71.9^\circ\)

ABCDEFGH is a cuboid with AB = 9 cm, BC = 4 cm and AE = 5 cm. The diagonals AG and EC cross at O. Find the acute angle between AG and EC.

• \(AG^2 = 9^2 + 4^2 + 5^2 = 122\), so \(AO = OC = \tfrac12\sqrt{122}\) (the diagonals cross at their midpoints)
• \(AE^2 = 5² = 25\)
• \(\cos AOE = \dfrac{30.5 + 30.5 - 25}{2 \times 30.5} = 0.5902\)
• \(\angle AOE = 53.83^\circ\), which is acute

ABCDE is a pyramid. The base ABCD is a square of side 10 cm. Each sloping edge is 14 cm.

1. (a) Find angle EBC.

   • \(\cos \angle EBC = \dfrac{5}{14}\) (BM is half of BC, in the isosceles triangle EBC)
   • \(\angle EBC = 69.1^\circ\)

2. (b) Find angle EBD.

   • \(BD = \sqrt{10^2 + 10^2}\), so \(BN = 7.071\)
   • \(\cos \angle EBD = \dfrac{7.071}{14}\), \(\angle EBD = 59.7^\circ\)

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

   • \(h^2 = 14^2 - 7.071^2 = 146.0\)
   • \(h = 12.1\) cm

ABCDEFGH is a cuboid with AB = 8 cm, BC = 7 cm and AE = 4 cm. Find the angle between the diagonal AG and the base ABCD.

• \(AC^2 = 8^2 + 7^2 = 113\)
• \(\tan \theta = \dfrac{CG}{AC} = \dfrac{4}{\sqrt{113}}\)
• \(\theta = 20.6^\circ\)

ABCDEFGH is a cuboid with AB = 8 cm, BC = 4 cm and AE = 3 cm. Find the angle between the diagonal AG and the face CDHG.

• \(AD\) is perpendicular to the face CDHG, so the shadow of AG on the face is DG. The angle is at G.
• \(DG^2 = 8^2 + 3^2 = 73\)
• \(\tan \theta = \dfrac{AD}{DG} = \dfrac{4}{\sqrt{73}}\), \(\theta = 25.1^\circ\)

A wardrobe is a cuboid 150 cm wide, 60 cm deep and 90 cm high. A long curtain pole is put inside 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 pole that fits, which is the diagonal AG.

   • \(AG^2 = 150^2 + 60^2 + 90^2 = 34200\)
   • \(AG = 185\) cm

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

   • \(AC^2 = 150^2 + 60^2 = 26100\)
   • \(\tan \theta = \dfrac{90}{\sqrt{26100}}\)
   • \(\theta = 29.1^\circ\)

A model roof is a square-based pyramid. The square base has side 6 m and each sloping edge is 5.2 m long.

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

   • \(BN = \tfrac12 \times 6\sqrt{2} = 4.243\)
   • \(h^2 = 5.2^2 - 4.243^2 = 9.040\)
   • \(h = 3.01\) m

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

   • \(\cos \theta = \dfrac{BN}{EB} = \dfrac{4.243}{5.2}\)
   • \(\theta = 35.3^\circ\)

A paperweight is a pyramid ABCD. The base ABC is an equilateral triangle of side 12 cm. The point D is directly above M, the centre of the base, and AD = 15 cm.

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

   • \(\cos 30^\circ = \dfrac{6}{AM}\), so \(AM = 6.928\)
   • \(\cos \theta = \dfrac{6.928}{15}\), \(\theta = 62.5^\circ\)

2. (b) Find the height DM.

   • \(DM^2 = 15^2 - 6.928^2 = 177.0\)
   • \(DM = 13.3\) cm

A parcel has a square base ABCD of side 8 cm. The diagonal AG makes an angle of 42° with the base. Find the height of the parcel, h.

• \(AC^2 = 8^2 + 8^2 = 128\)
• \(\tan 42^\circ = \dfrac{h}{\sqrt{128}}\)
• \(h = \sqrt{128} \tan 42^\circ = 10.2\) cm