Exercises: sin, cos and tan

Sketch

First thoughts

The triangle $△AKF$ has a right angle at $F$.

Therefore, the formulas for sin, cos and tan apply to it.

To know which of the formulas you should use, first determine,

• which side is the hypotenuse in the triangle $△AKF$, and

• which side is the adjacent cathetus to $\alpha$,

• and which one is the opposite cathetus to $\alpha$.

Determining hypotenuse, adjacent and opposite cathetus

• Hypotenuse: side [AK] with $\overline{\mathrm{AK}}={?}_{}$

• adjacent cathetus to $\alpha$: side [FA] with $\overline{\mathrm{AK}}={?}_{}$

• opposite cathetus to $\alpha$: side [FK] with $\overline{\mathrm{AK}}={?}_{}$

The formulas for sin, cos and tan are:

$\sin \alpha =\frac{\text{oppositecathetus}}{\text{hypotenuse}}$

$\cos \alpha =\frac{\text{adjacent cathetus}}{\text{hypotenuse}}$

$\tan \alpha =\frac{\text{opposite cathetus}}{\text{adjacent cathetus}}$

Since the hypotenuse is not given, it should preferably not appear in the formula you use.

$\to$ Solve the problem with the tangent.

Solution

$\tan \alpha =\frac{\text{opposite cathetus to }\alpha }{\text{adjacent cathetus to}\alpha }$

Plug in the lengths from the task into this formula.

$\tan \alpha =\frac{\overline{\mathrm{FK}}}{\overline{\mathrm{FA}}}$

Now insert the numerical values for the lengths of the distances and calculate the tangent of the angle.

$\tan \alpha =\frac{1.55\mathrm{m}}{12\mathrm{m}}=\frac{31}{240}\approx 0.129$

To obtain the angle $\alpha$, apply the inverse function ${\tan }^{-1}$ with the calculator.

$\alpha ={\tan }^{-1}(\frac{31}{240})\approx {7.4}^{\circ }$

⇒ The sun's rays fall on the road at an angle of ${7.4}^{\circ }$.