Question

circular railroad curves are designated by the degree of curvature, the central angle subtended by a chord of 100 ft. suppose a portion of track has a curvature of 28.0°. find (a) the radius of the curve. (round to the nearest integer as needed.) (a) the radius of the curve is 207 ft. (b) the length of the arc determined by the 100 - ft chord. (round to the nearest integer as needed.) (b) the length of the arc determined by the 100 - ft chord is 101 ft. (c) the area bounded by the arc and the 100 - ft chord is □ ft² (round to the nearest integer as needed.)

Explanation:

Step1: Recall arc - chord - radius relationship

Let the radius of the circle be $r$, the length of the chord be $l = 100$ ft and the central angle be $\theta=28.0^{\circ}=\frac{28\pi}{180}$ radians. We know that the length of the chord $l = 2r\sin(\frac{\theta}{2})$.

Step2: Solve for radius $r$

From $l = 2r\sin(\frac{\theta}{2})$, we can express $r=\frac{l}{2\sin(\frac{\theta}{2})}$. Substituting $l = 100$ and $\theta = 28^{\circ}$, so $\frac{\theta}{2}=14^{\circ}$ and $\sin(14^{\circ})\approx0.2419$. Then $r=\frac{100}{2\times0.2419}=\frac{100}{0.4838}\approx207$ ft.

Step3: Find the length of the arc

The formula for the length of an arc $s$ is $s = r\theta$, where $\theta$ is in radians. We have $r\approx207$ ft and $\theta=\frac{28\pi}{180}$ radians. So $s=207\times\frac{28\pi}{180}\approx207\times0.4887\approx101$ ft.

Step4: Find the area of the segment

The area of the sector $A_{sector}=\frac{1}{2}r^{2}\theta$ and the area of the triangle formed by the chord and the radii $A_{triangle}=\frac{1}{2}r^{2}\sin\theta$. The area of the segment $A = A_{sector}-A_{triangle}=\frac{1}{2}r^{2}(\theta - \sin\theta)$. Substituting $r = 207$ ft and $\theta=\frac{28\pi}{180}$ radians. $\sin\theta=\sin(28^{\circ})\approx0.4695$ and $\theta=\frac{28\pi}{180}\approx0.4887$. $A=\frac{1}{2}\times(207)^{2}(0.4887 - 0.4695)=\frac{1}{2}\times42849\times0.0192\approx411$ ft²

Answer:

(a) 207 ft
(b) 101 ft
(c) 411 ft²