contents...1 of 43 © boardworks ltd 2008 a a a a a a 7.4 finding angles contents s3 trigonometry...
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7.4 Finding angles
Contents
S3 Trigonometry
S3.5 Angles of elevation and depression
S3.6 Trigonometry in 3-D
S3.2 The three trigonometric ratios
S3.1 Right-angled triangles
S3.3 Finding side lengths
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The inverse of sin
sin θ = 0.5, what is the value of θ?
To work this out use the sin–1 key on the calculator.
sin–1 0.5 = 30°
sin–1 is the inverse of sin. It is sometimes called arcsin.
30° 0.5
sin
sin–1
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The inverse of sin
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The inverse of cos
Cos θ = 0.5, what is the value of θ?
To work this out use the cos–1 key on the calculator.
cos–1 0.5 = 60°
Cos–1 is the inverse of cos. It is sometimes called arccos.
60° 0.5
cos
cos–1
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The inverse of tan
tan θ = 1, what is the value of θ?
To work this out use the tan–1 key on the calculator.
tan–1 1 = 45°
tan–1 is the inverse of tan. It is sometimes called arctan.
45° 1
tan
tan–1
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4 steps we need to follow:
Step 1 Find which two sides we knowout of
Opposite, Adjacent and Hypotenuse.
Step 2 Use SOHCAHTOA to decide which one of
Sine, Cosine or Tangent ratio to use in this
question.
Step 3 For Sine calculate Opposite/Hypotenuse, for
Cosine calculate Adjacent/Hypotenuse or for
Tangent calculate Opposite/Adjacent.
Step 4 Find the angle from your calculator, using
one of sin-1, cos-1 or tan-1
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Finding angles
We are given the lengths of the sides opposite and adjacent to
the angle, so we use:
tan θ =opposite
adjacent
tan θ =8
5
= 57.99° (to 2 d.p.)
θ
5 cm
8 cm
θ = tan–1 (8 ÷ 5)
Find θ to 2 decimal places.
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Finding angles
We are given the lengths of the sides opposite and
hypotenuse to the angle, so we use:
sin θ =opposite
hypotenuse sin θ =2.5
5
= 30.00° (to 2 d.p.)
θ = sin–1 ( 2.5÷ 5)
Find θ to 2 decimal places.
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Finding angles
We are given the lengths of the sides opposite and adjacent to
the angle, so we use:
tan θ =opposite
adjacenttan θ =
300
400
= 36.87° (to 2 d.p.)
θ = tan–1 (300 ÷ 400)
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Finding angles
We are given the lengths of the sides hypotenuse and
adjacent to the angle, so we use:
cos θ =adjacent
hypotenusecos θ =
6750
8100
= 33.56° (to 2 d.p.)
θ = cos–1 (6750 ÷ 8100)
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Finding angles
We are given the lengths of the sides opposite and
hypotenuse to the angle, so we use:
sin θ =opposite
hypotenuse sin θ =18.88
30
= 39.00° (to 2 d.p.)
θ = sin–1 ( 18.88÷ 30)