Question

1. garry is making a temporary ramp from the ground to the top of his patio so that he can roll a cart up it to move furniture in. the surface of the patio is 3 feet 9 inches above the ground and garry wants the ramp to have a slope of no more than 0.375.

a) what is the shortest horizontal distance he can have from the base of the patio to the base of the ramp?
b) what is the shortest length the surface of the ramp can be?
c) what will be the angle of elevation of the ramp?

Explanation:

Step1: Convert height to inches

1 foot = 12 inches, so 3 feet 9 inches = 3×12 + 9=45 inches.

Step2: Use slope formula for part a

The slope formula is $m=\frac{y}{x}$, where $m$ is the slope, $y$ is the vertical - distance, and $x$ is the horizontal distance. Given $m = 0.375=\frac{3}{8}$ and $y = 45$ inches. We can solve for $x$: $x=\frac{y}{m}$. Substituting the values, $x=\frac{45}{\frac{3}{8}}=45\times\frac{8}{3}=120$ inches. Convert back to feet: $120\div12 = 10$ feet.

Step3: Use Pythagorean theorem for part b

The vertical distance $a = 45$ inches and the horizontal distance $b = 120$ inches. By the Pythagorean theorem $c=\sqrt{a^{2}+b^{2}}$, where $c$ is the length of the ramp. $c=\sqrt{45^{2}+120^{2}}=\sqrt{2025 + 14400}=\sqrt{16425}=128.16$ inches. Convert to feet: $128.16\div12\approx10.68$ feet.

Step4: Use tangent function for part c

The tangent of the angle of elevation $\theta$ is $\tan\theta=\frac{y}{x}$. Since $y = 45$ inches and $x = 120$ inches, $\tan\theta=\frac{45}{120}=0.375$. Then $\theta=\arctan(0.375)\approx20.56^{\circ}$.

Answer:

a) 10 feet
b) Approximately 10.68 feet
c) Approximately $20.56^{\circ}$