IB AA · Unit 2 · 5C Applications of Trigonometry
Find the angle between a sloping edge and the base, and the angle between a face and the base.
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Apex above the center
O is the center of the base and M is the apex. OM is the perpendicular height.
The angle between a sloping edge and the base is the angle at the base corner, in the right triangle from the corner to O to M.
Square base
Half the diagonal of the base is 12 × side × .
tan(edge angle) = heighthalf diagonal
tan(face angle) = heighthalf the side
The face angle uses the midpoint of a base edge, not the corner.
A cone
The angle between the sloping edge and the base uses tan θ = heightradius.
O is the center. The edge angle uses a corner. The face angle uses the midpoint of a side.
Half the base diagonal = 12 × 10 ×
Half the base diagonal = 7.07 cm
tan θ = heighthalf diagonal
tan θ = 127.07
θ = 59.5°
Answer: 59.5°
Half the base diagonal = 12 × 8 ×
Half the base diagonal = 5.66 cm
tan θ = heighthalf diagonal
tan θ = 95.66
θ = 57.8°
Answer: 57.8°
Half the base diagonal = 12 × 6 ×
Half the base diagonal = 4.24 cm
tan θ = heighthalf diagonal
tan θ = 84.24
θ = 62.1°
Answer: 62.1°
Distance from the center to the midpoint of a side = 12 × 14
That distance = 7 cm
tan θ = heightthat distance
tan θ = 107
θ = 55.0°
Answer: 55.0°
Distance from the center to the midpoint of a side = 12 × 12
That distance = 6 cm
tan θ = heightthat distance
tan θ = 56
θ = 39.8°
Answer: 39.8°
Half the base diagonal = 12 × 16 ×
Half the base diagonal = 11.3 cm
tan θ = heighthalf diagonal
tan θ = 1511.3
θ = 53.0°
Answer: 53.0°
For a cone, the right triangle uses the radius and the height.
tan θ = heightradius
tan θ = 125
θ = 67.4°
Answer: 67.4°
tan θ = heightradius
tan θ = 74
θ = 60.3°
Answer: 60.3°
tan θ = heightradius
tan θ = 86
θ = 53.1°
Answer: 53.1°
tan θ = heightradius
tan θ = 129
θ = 53.1°
Answer: 53.1°
Half diagonal =
Half diagonal = 5.00 cm
tan θ = 75.00
θ = 54.5°
Answer: 54.5°
The face along the 8 cm side is 3 cm from the center.
tan θ = 73
θ = 66.8°
Answer: 66.8°
Rearrange tangent to find the height or the base.
Half diagonal = 12 × 12 ×
Half diagonal = 8.49 cm
tan 58° = h8.49
h = 8.49 × tan 58°
h = 13.6 cm
Answer: 13.6 cm
tan 62° = h5
h = 5 × tan 62°
h = 9.40 cm
Answer: 9.40 cm
Distance from the center of an equilateral triangle to a vertex = 9
That distance = 5.20 cm
tan θ = 85.20
θ = 57.0°
Answer: 57.0°
Distance from the center of an equilateral triangle to a vertex = 6
That distance = 3.46 cm
tan θ = 113.46
θ = 72.5°
Answer: 72.5°
tan 48° = 9half side
half side = 9tan 48°
half side = 8.10 cm
side = 16.2 cm
Answer: 16.2 cm
tan 40° = 15r
r = 15tan 40°
r = 17.9 cm
sin 40° = 15l
l = 15sin 40°
l = 23.3 cm
Answer: r = 17.9 cm, l = 23.3 cm
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