Worked examples of the inverses

The inverse trigonometric functions crop up wherever an angle must be computed from a ratio. Three typical examples.

Example 1 — Angle of elevation

An observer sees the top of a tower of height $h=30$ m, from a horizontal distance $d=50$ m. The angle of elevation $\theta$ with respect to the ground satisfies: $\tan\theta = \frac{h}{d} = \frac{30}{50} = 0{,}6 \;\Rightarrow\; \theta = \arctan(0{,}6) \approx 30{,}96°.$ At distance $d=10$ m: $\theta = \arctan 3 \approx 71{,}57°$. As one approaches the tower, the angle grows towards $90°$ (vertical asymptote of $\tan$).

Example 2 — Angle between two lines

Given $r: y = 2x+1$ and $s: y = -\tfrac12 x + 3$, with gradients $m_1=2$, $m_2=-1/2$. Since $m_1 m_2 = -1$, the lines are perpendicular (angle $90°$). In general the angle $\alpha$ between two lines is obtained from $\tan\alpha = \frac{m_1 - m_2}{1+m_1 m_2},\qquad \alpha = \arctan\!\left(\frac{m_1-m_2}{1+m_1 m_2}\right).$ If $1+m_1m_2 = 0$ the denominator vanishes and the angle is $\pi/2$ (perpendicularity).

Example 3 — Refraction (an outline)

Snell’s law relates the angle of incidence $\theta_i$ and of refraction $\theta_r$ to the index $n$: $\sin\theta_i = n\sin\theta_r$. Knowing $\theta_i$ and $n$, the refracted angle is $\theta_r = \arcsin\!\left(\sin\theta_i/n\right)$. For water-air $n=1{,}33$ and $\theta_i=45°$: $\theta_r = \arcsin(0{,}531) \approx 32{,}1°$.

Links

Topics: Trigonometry
Concepts: Arcsine · Arctangent · Gradient · Line
Functions: Arcsine · Arctangent · Inverse trigonometric functions
Skills: Analytic geometry · Modelling · Using formulae