AI SL · 12.10
0 of 60 skills done
Angles in 3D shapes
12.10.5
Screenshot and video to come
Screenshot and video to come
Choose the angle you want. The right-angled triangle that holds it is shaded in the cuboid and drawn flat beside it. Change the three edges and watch the working.
Move a slider to change the cuboid.
ABCDEFGH is a cuboid with AB = 9 cm, BC = 4 cm and AE = 6 cm. Find the size of angle HAG.
Have a go first. Then open the working one line at a time.
\(HG \parallel AB\)
AB is perpendicular to the face ADHE, so HG is perpendicular to AH. Triangle AHG has its right angle at H.
\(AH^2 = 4^2 + 6^2 = 52\)
Pythagoras in the flat face first. Keep AH as a surd.
\(\tan \angle HAG = \dfrac{HG}{AH} = \dfrac{9}{\sqrt{52}}\)
HG is opposite the angle and AH is adjacent. HG is as long as AB.
From the calculator: tan⁻¹(9/√52) = 51.3°
Check the calculator is in degrees.
ABCDEFGH is a cuboid with AB = 9 cm, BC = 5 cm and AE = 4 cm. The diagonals AG and EC cross at O. Find the acute angle between AG and EC.
Have a go first. Then open the working one line at a time.
\(AG^2 = 9^2 + 5^2 + 4^2 = 122\), so \(AO = OC = \tfrac12\sqrt{122}\) (the diagonals cross at their midpoints)
Both diagonals have the same length and cross at O. That gives an isosceles triangle with two equal sides.
\(AC^2 = 9² + 5² = 106\)
AC is the diagonal of the base.
\(\cos AOC = \dfrac{30.5 + 30.5 - 106}{2 \times 30.5} = -0.7377\)
The triangle is not right-angled, so use the cosine rule. A negative cosine means an obtuse angle.
\(137.5^\circ \text{ is obtuse: } 180^\circ - 137.5^\circ = 42.5^\circ\)
Two lines make two angles that add to 180°. The angle between the lines is the acute one.
ABCDE is a pyramid. The base ABCD is a square of side 8 cm. Each sloping edge is 11 cm. Find the angle EBC.
Have a go first. Then open the working one line at a time.
\(EB = EC = 11\)
The sloping edges are equal, so the perpendicular from E meets BC at its midpoint M. Then BM = 4.
\(\cos \angle EBC = \dfrac{BM}{EB} = \dfrac{4}{11}\)
BM is adjacent to the angle, and EB is the hypotenuse.
From the calculator: cos⁻¹(4/11) = 68.7°
Give 3 significant figures.
ABCDE is a pyramid. The base ABCD is a square of side 8 cm. Each sloping edge is 11 cm. Find the angle EBD.
Have a go first. Then open the working one line at a time.
\(BD = \sqrt{8^2 + 8^2} = 8\sqrt{2}\)
Pythagoras in the square base.
\(BN = \tfrac12 BD = 4\sqrt{2} = 5.657\)
The apex is directly above the centre N, which is the middle of BD.
\(\cos \angle EBD = \dfrac{BN}{EB} = \dfrac{5.657}{11}\)
Triangle ENB has a right angle at N. BN is adjacent and EB is the hypotenuse.
From the calculator: cos⁻¹(5.657/11) = 59.1°
Give 3 significant figures.
ABCDE is a pyramid. The base ABCD is a square of side 8 cm. Each sloping edge is 11 cm. Find the height of the pyramid.
Have a go first. Then open the working one line at a time.
\(EN^2 + BN^2 = EB^2\)
EN is the height. The right angle is at N, the centre of the base.
\(BN^2 = (4\sqrt{2})^2 = 32\)
BN is half the diagonal of the base.
\(h^2 = 11^2 - 32 = 89\), \(h = 9.43\)
Take the positive square root. The height is in centimetres.
ABCDEFGH is a cuboid with AB = 5 cm, BC = 3 cm and AE = 7 cm. Find the angle between the diagonal AG and the face CDHG.
Have a go first. Then open the working one line at a time.
\(AD \perp \text{face } CDHG\)
AD is perpendicular to the face, so D is the foot of the perpendicular from A. The shadow of AG on the face is DG. The angle is at G, between AG and DG.
\(DG^2 = 5^2 + 7^2 = 74\)
Pythagoras in the face CDHG. AD is as long as BC, which is 3.
\(\tan \theta = \dfrac{AD}{DG} = \dfrac{3}{\sqrt{74}}\)
AD is opposite the angle at G. DG is adjacent.
From the calculator: tan⁻¹(3/√74) = 19.2°
Give 3 significant figures.
ABCDEFGH is a cuboid with AB = 10 cm, BC = 6 cm and AE = 7 cm. Find the angle between the diagonal AG and the base ABCD.
Have a go first. Then open the working one line at a time.
\(GC \perp \text{base}\)
C is the foot of the perpendicular from G. The shadow of AG on the base is AC. The angle is at A.
\(AC^2 = 10^2 + 6^2 = 136\)
Pythagoras in the base.
\(\tan \theta = \dfrac{CG}{AC} = \dfrac{7}{\sqrt{136}}\)
CG is the height of the cuboid.
From the calculator: tan⁻¹(7/√136) = 31.0°
Give 3 significant figures.
ABCD is a pyramid. The base ABC is an equilateral triangle of side 9 cm. The point D is directly above M, the centre of the base, and AD = 12 cm. Find the angle between AD and the base.
Have a go first. Then open the working one line at a time.
\(DM \perp \text{base}\)
M is the foot of the perpendicular from D. The shadow of AD on the base is AM. The angle is at A.
\(\cos 30^\circ = \dfrac{4.5}{AM}\), so \(AM = 5.196\)
AM bisects the angle at A, so it makes 30° with AB. Look at the right-angled triangle with half of AB.
\(\cos \theta = \dfrac{AM}{AD} = \dfrac{5.196}{12}\)
Triangle DMA has its right angle at M. AM is adjacent, AD is the hypotenuse.
From the calculator: cos⁻¹(5.196/12) = 64.3°
Give 3 significant figures.