Question

1. lombard street in san francisco, california, is paved with red bricks and is a switchback road with a speed limit of 8 km/h because of its extreme grade. the angle of elevation of lombard street in one section reaches 15°. determine the grade of this section of road to the nearest percent. step 1: find slope step 2: find percent grade

Explanation:

Step1: Find slope

The slope \(m\) of a line with angle of elevation \(\theta\) is given by \(m = \tan\theta\). Here \(\theta = 15^{\circ}\), so \(m=\tan(15^{\circ})\).
Using the formula \(\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}\), and \(15^{\circ}=45^{\circ}-30^{\circ}\), \(\tan(45^{\circ}) = 1\), \(\tan(30^{\circ})=\frac{\sqrt{3}}{3}\).
\(\tan(15^{\circ})=\tan(45^{\circ}-30^{\circ})=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=\frac{3-\sqrt{3}}{3 + \sqrt{3}}\).
Rationalizing the denominator: \(\frac{(3-\sqrt{3})(3-\sqrt{3})}{(3 + \sqrt{3})(3-\sqrt{3})}=\frac{9-6\sqrt{3}+ 3}{9-3}=\frac{12-6\sqrt{3}}{6}=2-\sqrt{3}\approx2 - 1.732=0.268\).

Step2: Find percent grade

The grade of a road is given by \(Grade=m\times100\%\).
Since \(m\approx0.268\), then \(Grade = 0.268\times100\%=26.8\%\approx27\%\).

Answer:

\(27\%\)