Example 9 from Algebra and Trigonometry, 9.3 Double-Angle, Half-Angle, and Reduction Formulas

Finding the Measurement of a Half Angle

Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of $θ$ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If $\tan θ = \frac{5}{3}$ for higher-level competition, what is the measurement of the angle for novice competition?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Since the angle for novice competition measures half the steepness of the angle for the high level competition, and $\tan θ = \frac{5}{3}$ for high competition, we can find $\cos θ$ from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See Figure 4.

$\begin{matrix} 3^{2} + 5^{2} & = & 34 \\ c & = & \sqrt{34} \end{matrix}$

Figure 4

We see that $\cos θ = \frac{3}{\sqrt{34}} = \frac{3 \sqrt{34}}{34} .$ We can use the half-angle formula for tangent: $\tan \frac{θ}{2} = \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} .$ Since $\tan θ$ is in the first quadrant, so is $\tan \frac{θ}{2} .$
$\begin{matrix} \tan \frac{θ}{2} & = & \sqrt{\frac{1 - \frac{3 \sqrt{34}}{34}}{1 + \frac{3 \sqrt{34}}{34}}} \\ = & \sqrt{\frac{\frac{34 - 3 \sqrt{34}}{34}}{\frac{34 + 3 \sqrt{34}}{34}}} \\ = & \sqrt{\frac{34 - 3 \sqrt{34}}{34 + 3 \sqrt{34}}} \\ \approx & 0.57 \end{matrix}$
We can take the inverse tangent to find the angle: $\tan^{- 1} (0.57) \approx 29.7° .$ So the angle of the ramp for novice competition is $\approx 29.7° .$

How did it go?

Solution

Solution