The car and the observer

In the figure, $J$ represents the position of James and $C$ represents the car which moves through a distance of $x$ first from $C$ to $A$ and then from $A$ to $B$ . With $a$ and $b$ represent the angles of rotation of the neck of James.

In triangle $ACJ$ , $\begin{aligned} \sin(a)&=&\frac{AC}{AJ} \\ & =& \frac{x}{AJ} \quad (\text{as } AC=x) \qquad \qquad \tag1 \end{aligned}$

In triangle $ADJ$ , $\sin(b)=\frac{AD}{AJ} \qquad \qquad \tag2$

In triangle $ADB$ , $AD<AB \quad (\text{as }AB \text{ is the hypotenuse})$

or $AD < x\qquad \qquad \tag 3$

as $AB = x$ .

From $(1), (2), (3)$ , we get

$\sin a > \sin b .$

Since $0<a< \frac\pi2$ and $0 < b < \frac\pi2$ , then $\boxed{a>b}$ .

By interpreting the problem statement, we can construct the diagram as shown above, where $x$ and $y$ are positive numbers.

We have $\tan a = \dfrac xy$ and $\tan (a+b) = \dfrac{x+x}y = \dfrac{2x}y$ .

By applying the compound angle formula for tangent, $\tan(M-N) = \dfrac{\tan M - \tan N}{1 + \tan M \tan N }$ , we can determine $\tan b$ ,

$\begin{aligned} \tan b = \tan ((a+b) - a) &=& \dfrac{ \tan (a+b) - \tan a}{1 + \tan(a+b) \tan a} \\ &=& \dfrac{\frac{2x}y - \frac xy }{1 + \frac{2x}y \cdot \frac xy } \\ &=& \dfrac{ \frac xy }{1 + \frac{2x^2}{y^2} } \\ &=& \dfrac{ xy }{2x^2 + y^2 } \\ \end{aligned}$

Since $\tan a = \dfrac xy = \dfrac{xy}{y^2} > \dfrac{xy}{2x^2 + y^2} = \tan b$ , and because $a$ and $b$ are in the interval $\left( 0 ,\frac\pi2 \right)$ , then $\boxed{a > b}$ because the graph $f(z)= \tan z$ is an increasing function in the first quadrant.