Questions: A trigonometry student stands 15 meters from a building and measures the angle of elevation to the top of the building as 30°. How accurate does her angle measurement have to be if she wants her propagated percentage error in estimating the height of the building to be no more than 4%? Round any intermediate calculations to no less than six decimal places, and round your final answer to two decimal places.

Solution Steps

Let h be the height of the building. We have $\tan(30^{\circ}) = \frac{h}{15}$, so $h = 15\tan(30^{\circ}) = 15\frac{1}{\sqrt{3}} = \frac{15\sqrt{3}}{3} = 5\sqrt{3} \approx 8.660254$ m.

We have $h = 15 \tan(\theta)$. Then $dh = 15 \sec^2(\theta) d\theta$. We are given that $\theta = 30^\circ = \frac{\pi}{6}$ radians. Then $dh = 15 \sec^2(30^\circ) d\theta = 15 (\frac{2}{\sqrt{3}})^2 d\theta = 15 (\frac{4}{3}) d\theta = 20 d\theta$.

We are given that the propagated percentage error in estimating the height is no more than 4%. This means that $\frac{|dh|}{h} \le 0.04$. Substituting $dh = 20 d\theta$ and $h = 5\sqrt{3}$, we get $\frac{|20 d\theta|}{5\sqrt{3}} \le 0.04$ $|d\theta| \le \frac{0.04 \times 5\sqrt{3}}{20} = 0.01\sqrt{3} \approx 0.01732$ radians. In degrees, $|d\theta| \le 0.01732 \times \frac{180}{\pi} \approx 0.9922^\circ$.

Final Answer: The angle measurement must be accurate to within $0.99^{\circ}$.