Linear Equations II

Hong Kong

Stage 4 - Stage 5

Lesson

The angle between two lines in the Cartesian plane can be found from their gradients using some trigonometry involving the tangent function.

Another approach, using the cosine function and ideas about vectors, is available if pairs of points on the two lines are known. However, in this chapter, we consider only the case where the gradients are known.

We understand the angle between two lines to be the acute angle.

First, recall some basic ideas. The gradient or slope of a line is measured by the fraction $\frac{\text{rise}}{\text{run}}$riserun as in the following diagram. But, this is the same as $\tan\theta$`t``a``n``θ`. So there is a connection between *angle *and *gradient*.

In the following diagram, there are two lines whose angles of inclination to the horizontal axis are known, $\alpha$`α` and $\beta$`β` respectively. We can deduce the angle $\theta$`θ` between the two lines with some simple geometry.

There is a triangle in the diagram with angles $\beta$`β`, $\theta$`θ` and $180^\circ-\alpha$180°−`α`. These sum to $180^\circ$180°. It follows that

$\theta=\alpha-\beta$`θ`=`α`−`β`.

Now, gradients are the *tangents *of the angles of inclination. So, we write

$\tan\theta=\tan\left(\alpha-\beta\right)$`t``a``n``θ`=`t``a``n`(`α`−`β`)

and therefore,

$\tan\theta=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$`t``a``n``θ`=`t``a``n``α`−`t``a``n``β`1+`t``a``n``α``t``a``n``β`

using a previously established angle sum formula for the tangent function. This formula tells us that we can find the tangent of the angle between two lines if we know the gradients of the two lines.

Suppose we have two lines $y=2x+c$`y`=2`x`+`c` and $y=3x+d$`y`=3`x`+`d`. Their gradients are respectively $2$2 and $3$3. We wish to find the tangent of the angle between the two lines. (Notice that the intercepts $c$`c` and $d$`d` are not needed.)

We can put $\tan\alpha=3$`t``a``n``α`=3 and $\tan\beta=2$`t``a``n``β`=2. We require $\tan\theta$`t``a``n``θ` where $\theta$`θ` is the angle between the two lines.

According to the formula, $\tan\theta=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=\frac{3-2}{1+3\times2}=\frac{1}{7}$`t``a``n``θ`=`t``a``n``α`−`t``a``n``β`1+`t``a``n``α``t``a``n``β`=3−21+3×2=17

The angle between the lines is found using the inverse tangent function.

If $\tan\theta=\frac{1}{7}$`t``a``n``θ`=17, then $\theta=\arctan\frac{1}{7}\approx8.13^\circ$`θ`=`a``r``c``t``a``n`17≈8.13°.

If it should happen that the two lines are parallel, then $\tan\alpha=\tan\beta$`t``a``n``α`=`t``a``n``β` and the numerator in the formula is zero making the angle of intersection zero.

If one of the lines is parallel to the vertical axis, the formula cannot be used because the angle of inclination for this line would be $90^\circ$90° and $\tan90^\circ$`t``a``n`90° is undefined. The angle between the two lines can be worked out easily from the angle of inclination of the other line in this case.

The graphs of $y=x^2$`y`=`x`2 and $y=\sqrt{x}$`y`=√`x` are shown in the diagram below. The two curves intersect at the points $(0,0)$(0,0) and $(1,1)$(1,1). We define the angle of intersection in each case to be the angle of intersection of the tangent lines at those points.

At the origin, the curves intersect at a right angle. To find the angle of intersection at the point $(1,1)$(1,1), we need to know the gradients of the tangents at that point.

Using some calculus, we find that the gradient of $y=x^2$`y`=`x`2 at $(1,1)$(1,1) is $2$2, and the gradient of $y=\sqrt{x}$`y`=√`x` at that point is $\frac{1}{2}$12.

If we call the angle between the two tangent lines $\theta$`θ`, we have

$\tan\theta=\frac{2-\frac{1}{2}}{1+2\times\frac{1}{2}}=\frac{3}{4}$`t``a``n``θ`=2−121+2×12=34.

Then,

$\theta=\arctan\frac{3}{4}\approx36.9^\circ$`θ`=`a``r``c``t``a``n`34≈36.9°.

Find an expression for $\tan\theta$`t``a``n``θ`, where $\theta$`θ` is the acute angle between the two lines in the diagram.

Solve for the acute angle $\theta$`θ` between the line with slope $\frac{5}{9}$59 and the line with slope $\frac{7}{2}$72.

Give your answer in degrees.

Solve for the acute angle $\theta$`θ` between the lines $x+2y=-4$`x`+2`y`=−4 and $x-5y=-8$`x`−5`y`=−8.

Give your answer to the nearest tenth of a degree.