IB AA · Unit 2 · 5C Applications of Trigonometry
Find a diagonal of a box and the angle it makes with an edge or with the base.
Label the box
ABCD is the bottom rectangle and EFGH is the top. E is above A, F above B, G above C, and H above D.
AB is the length, BC is the width, and AE is the height. Diagonal AC crosses the base. AG is a space diagonal.
A right triangle on a face
A face diagonal is on that face.
The angle between that diagonal and an edge uses tangent: opposite over adjacent.
The space diagonal
AG =
The angle between AG and the base uses tan θ = heightbase diagonal.
ABCD is the base. AC joins opposite corners of the base.
AC =
AC =
AC =
AC = 10 cm
tan (angle between AC and AB) = BCAB
tan θ = 68
θ = 36.9°
Answer: AC = 10 cm, θ = 36.9°
AC =
AC = 15 cm
tan θ = 129
θ = 53.1°
Answer: AC = 15 cm, θ = 53.1°
AC =
AC = 25 cm
tan θ = 247
θ = 73.7°
Answer: AC = 25 cm, θ = 73.7°
AC =
AC = 17 cm
tan θ = 815
θ = 28.1°
Answer: AC = 17 cm, θ = 28.1°
AC =
AC = 29 cm
tan θ = 2120
θ = 46.4°
Answer: AC = 29 cm, θ = 46.4°
AC =
AC = 13 cm
tan θ = 125
θ = 67.4°
Answer: AC = 13 cm, θ = 67.4°
AG runs from a bottom corner to the opposite top corner.
AC =
AC = 10.0 cm
AG =
AG =
AG = 11.2 cm
tan θ = AEAC
tan θ = 510.0
θ = 26.6°
Answer: AG = 11.2 cm, θ = 26.6°
AC =
AC = 15.0 cm
AG =
AG =
AG = 17.0 cm
tan θ = AEAC
tan θ = 815.0
θ = 28.1°
Answer: AG = 17.0 cm, θ = 28.1°
AC =
AC = 8.06 cm
AG =
AG =
AG = 9.00 cm
tan θ = AEAC
tan θ = 48.06
θ = 26.4°
Answer: AG = 9.00 cm, θ = 26.4°
AC =
AC = 14.1 cm
AG =
AG =
AG = 15.8 cm
tan θ = AEAC
tan θ = 714.1
θ = 26.3°
Answer: AG = 15.8 cm, θ = 26.3°
AC =
AC = 25.0 cm
AG =
AG =
AG = 27.7 cm
tan θ = AEAC
tan θ = 1225.0
θ = 25.6°
Answer: AG = 27.7 cm, θ = 25.6°
AG =
AG = 11.2 cm
cos θ = ABAG
cos θ = 611.2
θ = 57.5°
Answer: 57.5°
Find the third edge of the right triangle before you use tangent or cosine.
AC =
AC = 8.06 cm
AG =
AG =
AG = 12.1 cm
tan θ = AEAC
tan θ = 98.06
θ = 48.1°
Answer: AG = 12.1 cm, θ = 48.1°
Adding the three face-diagonal equations counts every edge twice.
2 AG² = (10.0)² + (12.0)² + (10.8)²
AG = 13.5 cm
The base diagonal is the face diagonal of the base.
Height² = 12 × ((12.0)² + (10.8)² − (10.0)²)
Height = 9.00 cm
tan θ = 9.0010.0
θ = 42.0°
Answer: AG = 13.5 cm, θ = 42.0°
AC =
AC = 8.60 cm
AG =
AG =
AG = 11.7 cm
tan θ = AEAC
tan θ = 88.60
θ = 42.9°
AG =
AG = 11.7 cm
cos θ = ABAG
cos θ = 511.7
θ = 64.8°
Answer: with the base 42.9°, with AB 64.8°
AC =
AC = 13.0 cm
AG =
AG =
AG = 20.6 cm
tan θ = AEAC
tan θ = 1613.0
θ = 50.9°
Answer: AG = 20.6 cm, θ = 50.9°
AG =
AG = 25.0 cm
cos θ = ABAG
cos θ = 925.0
θ = 68.9°
Answer: 68.9°
AG =
AG = 14.0 cm
cos θ = ABAG
cos θ = 1114.0
θ = 38.0°
Answer: 38.0°
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