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Pythagoras and Trigonometry in 3D

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AQA Level 2 Further Maths practice questions, each with a hint and a full worked solution.

Have a pen, paper and a calculator handy, to work out your answers.

Question 12 marks

The diagonal joining opposite corners of a cube is 12 cm long.

Work out the length of one edge of the cube.

Give your answer as a surd in its simplest form.

Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

Hint

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

Worked solution
  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

Question 23 marks

ABCDEFGH is a cuboid.

AB=9 cm, \;BC=4 cm and \;CG=6 cm

Work out the angle between the diagonal AG and the base ABCD.

Give your answer to 1 decimal place.

Hint

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

Worked solution
  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

Question 3Challenge5 marks

VABCD is a pyramid with a horizontal square base ABCD of side 10 cm.

X is the centre of the base and V is vertically above X with \;VX=14 cm

M is the midpoint of the edge VC.

Work out the angle between the line BM and the base ABCD.

Give your answer to 1 decimal place.

Hint

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

Worked solution
  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

More on this topic: Pythagoras and Trigonometry in 3D worksheet with full solutions

All AQA Level 2 Further Maths practice questions

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