IB AA · Unit 2 · 5B Rules of Trigonometry
Use the gradient to find the angle a line makes with the x-axis, and the angle between two lines.
Angle with the positive x-axis
A line y = mx + c makes angle θ with the positive x-axis.
tan θ = m, so θ = tan−1(m).
A negative gradient still makes a positive-sized angle. Use the positive value of the gradient for that size.
Angle between two lines
tan θ = m2 − m11 + m1 m2
Use the positive size of that value. The result is the acute angle.
If the denominator is 0, the lines are perpendicular.
Intercepts
The y-intercept is the constant term.
The x-intercept is the x-value when y = 0.
The gradient is the tangent of the angle.
θ = tan−1(gradient)
θ = tan−1(3)
θ = 71.6°
Answer: 71.6°
θ = tan−1(gradient)
θ = tan−1(2)
θ = 63.4°
Answer: 63.4°
θ = tan−1(gradient)
θ = tan−1(12)
θ = 26.6°
Answer: 26.6°
θ = tan−1(gradient)
θ = tan−1(5)
θ = 78.7°
Answer: 78.7°
θ = tan−1(gradient)
θ = tan−1(−2)
θ = 63.4°
Answer: 63.4°
θ = tan−1(gradient)
θ = tan−1(34)
θ = 36.9°
Answer: 36.9°
Take the positive value so the angle you report is acute.
tan θ = m2 − m11 + m1 m2
tan θ = −1030
The denominator is 0, so the lines are perpendicular.
θ = 90°
Answer: 90°
tan θ = m2 − m11 + m1 m2
tan θ = −20
The denominator is 0, so the lines are perpendicular.
θ = 90°
Answer: 90°
tan θ = m2 − m11 + m1 m2
tan θ = −520
The denominator is 0, so the lines are perpendicular.
θ = 90°
Answer: 90°
tan θ = m2 − m11 + m1 m2
tan θ = −35
θ = 31.0°
Answer: 31.0°
tan θ = m2 − m11 + m1 m2
tan θ = −311
θ = 15.3°
Answer: 15.3°
tan θ = m2 − m11 + m1 m2
tan θ = 5−5
θ = 45°
Answer: 45°
Find where the line meets the axes before you use the gradient.
When x = 0, y = −6
y-intercept is (0, −6)
When y = 0,
0 = 2x − 6
2x = 6
x = 3
x-intercept is (3, 0)
θ = tan−1(2)
θ = 63.4°
Answer: intercepts (0, −6) and (3, 0), θ = 63.4°
Set the y-values equal.
2x − 4 = −x + 2
3x = 6
x = 2
y = 0
The lines meet at (2, 0), on the x-axis.
tan θ = m2 − m11 + m1 m2
tan θ = −3−1
θ = 71.6°
Answer: they meet at (2, 0), θ = 71.6°
tan θ = m2 − m11 + m1 m2
tan θ = −5−5
θ = 45°
Answer: 45°
tan θ = m2 − m11 + m1 m2
tan θ = −1740
The denominator is 0, so the lines are perpendicular.
θ = 90°
Answer: 90°
θ = tan−1(gradient)
θ = tan−1(52)
θ = 68.2°
Answer: 68.2°
tan θ = m2 − m11 + m1 m2
tan θ = −422
θ = 10.3°
Answer: 10.3°
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