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Topos Uranos

Summary

This lesson uses trigonometry to measure the angle formed by two lines in the plane. It defines the inclination of a line and proves that its tangent is the slope; with the formula for the tangent of a difference it obtains the tangent of the angle between two lines from their slopes, which contains the perpendicularity condition as a special case, and with the dot product it gives a second expression, also valid for vertical lines, whose agreement with the first is Lagrange’s identity. It then characterizes the bisectors of two lines as the locus of the points equidistant from both, proves that they are perpendicular and uses them to obtain the incenter and the inscribed circle of a triangle. The last section gathers applications: the height of an inaccessible point by means of the law of sines, simple harmonic motion as the shadow of uniform circular motion, the range of a projectile, which is greatest at 45∘45^\circ and the same for complementary angles, and the law of reflection as a property of the shortest path.

Learning objectives

  1. Compute the inclination of a line and the angle between two lines, from their slopes or from their direction vectors, and recognize parallelism and perpendicularity as the extreme cases.
  2. Obtain the equations of the bisectors of two lines as a locus of equidistant points, choose the internal bisector of a triangle and compute its incenter and its inscribed circle.
  3. Solve problems of heights and distances with angles of elevation and of depression, combining the law of sines with the sum and difference formulas.
  4. Describe simple harmonic motion, the range of a projectile and the reflection of light with trigonometric functions, and derive their properties from the identities of this unit.

The inclination of a line

In the Course I lesson on the equation of a line, the direction of a non-vertical line was measured by a number, its slope, and it was proved that two non-vertical lines are perpendicular if and only if the product of their slopes is −1-1. That condition recognizes one particular angle, the right angle, without measuring any angle. In the lesson on trigonometric ratios it was also seen that the tangent of an angle is the slope of the line containing its terminal side. If this is so, the slope is the tangent of an angle, and the angles between lines should be computable with trigonometry.

The angle of inclination of a line is the angle α\alpha, with 0≤α<π0 \le \alpha < \pi, that the positive half of the horizontal axis forms with the line, measured counterclockwise from the axis to the line. Horizontal lines have inclination 00, and vertical lines, inclination π2\frac{\pi}{2}.

More precisely, the parallel to the line through the origin contains the terminal side of some angle α\alpha, and also that of α+π\alpha + \pi, because the points P(α)P(\alpha) and P(α+π)P(\alpha + \pi) of the unit circle are opposite; exactly one of the angles of the form α+kπ\alpha + k\pi lies in [0,π)\ico{0}{\pi}, and that one is the inclination. Two lines are parallel or coincide if and only if they have the same inclination, because then, and only then, they have the same parallel through the origin.

LemmaInclination and slope

A line of inclination α\alpha is vertical if and only if α=π2\alpha = \frac{\pi}{2}; if it is not vertical, its slope is the tangent of its inclination. Conversely, the inclination of a line of slope mm is arctan⁡m\arctan m if m≥0m \ge 0, and π+arctan⁡m\pi + \arctan m if m<0m < 0.

m=tan⁡αm = \tan \alpha

Everything is read on the parallel to the line through the origin, which contains the point P(α)P(\alpha) of the unit circle. The figure uses the lines y=3x+1y = \sqrt{3} x + 1 and y=−xy = -x.

Proof

  1. O=(0,0)∈r0,P(α)=(cos⁡α,sin⁡α)∈r0\dato{p}{O = \left( 0, 0 \right) \in r_0, \quad P(\alpha) = \left( \cos \alpha, \sin \alpha \right) \in r_0}

    Let rr be a line of inclination α\alpha, and let r0r_0 be its parallel through the origin. By the definition of the inclination, r0r_0 contains the terminal side of α\alpha, and in particular the point P(α)P(\alpha); if rr is not vertical, r0r_0 has the same slope, because two parallel lines have the same slope, according to the Course I lesson on the equation of a line. In the figure, α=π3\alpha = \frac{\pi}{3}.

  2. 0≤α<π,cos⁡α=0↔α=π20 \leq \alpha < \pi, \quad \resaltar{\cos \alpha = 0} \leftrightarrow \alpha = \frac{\pi}{2}

    The line r0r_0 is vertical if and only if its two points have the same abscissa, that is, if cos⁡α=0\cos \alpha = 0. In [0,π)\ico{0}{\pi}, the only zero of the cosine is π2\frac{\pi}{2}; and rr is vertical exactly when its parallel is.

  3. m=sin⁡α−0cos⁡α−0=tan⁡αm = \frac{\sin \alpha - 0}{\cos \alpha - 0} = \dato{m}{\resaltar{\tan \alpha}}

    If rr is not vertical, its slope is that of r0r_0, which is the quotient of the difference of the ordinates by that of the abscissas of its two points.

  4. α={arctan⁡mm≥0π+arctan⁡mm<0\alpha = \begin{cases} \resaltar{\arctan m} & m \geq 0 \\ \resaltar{\pi + \arctan m} & m < 0 \end{cases}

    Conversely, let mm be the slope of rr. By the step above, the inclination α\alpha is an angle in [0,π)\ico{0}{\pi} whose tangent is mm. According to the lesson on trigonometric equations, the angles with tangent mm are the numbers arctan⁡m+kπ\arctan m + k\pi, and exactly one of them lies in [0,π)\ico{0}{\pi}: since arctan⁡m\arctan m lies in (−π2,π2)\ioo{-\frac{\pi}{2}}{\frac{\pi}{2}}, that angle is arctan⁡m\arctan m if m≥0m \ge 0, and π+arctan⁡m\pi + \arctan m if m<0m < 0. In the figure, the line y=−xy = -x has inclination π−π4=3π4\pi - \frac{\pi}{4} = \frac{3\pi}{4}.

For example, the line y=3x+1y = \sqrt{3} x + 1 has inclination arctan⁡3=π3\arctan \sqrt{3} = \frac{\pi}{3}, and the line y=−xy = -x, inclination π+arctan⁡(−1)=3π4\pi + \arctan \left( -1 \right) = \frac{3\pi}{4}. The sign of the slope thus distinguishes the lines that rise, with acute inclination, from those that fall, with obtuse inclination.