Chicago ave is parallel to Ontario Street. What is the relationship between angles 2 and 3?
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9514 1404 393
Answer:
A to D is about 2087 ft
Step-by-step explanation:
For the portion of the path of interest, the sine and cosine relations apply.
Sin = Opposite/Hypotenuse
Cos = Adjacent/Hypotenuse
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AB is the adjacent side to the angle marked 33°, with hypotenuse 975 ft.
cos(33°) = AB/(975 ft)
AB = (975 ft)cos(33°) ≈ 817.70 ft
BC is the opposite side in the same triangle.
sin(33°) = BC/(975 ft)
BC = (975 ft)sin(33°) ≈ 531.02 ft
CD is the hypotenuse of the right triangle BCD. The side BC that we know is opposite the given angle, so the sine relation applies.
sin(46°) = BC/CD
CD = BC/sin(46°)
CD = (531.02 ft)/sin(46°) ≈ 738.21 ft
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Now we know the segments of the path of interest:
A–D = AB +BC +CD
A–D = 817.70 ft + 531.02 ft + 738.21 ft = 2086.93 ft
A–D ≈ 2087 ft
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<em>Additional comments</em>
The attachment shows all of the segments computed working counterclockwise from A. The angles in triangle DEF are computed using the Law of Sines from sides DF and DE and angle DEF. Segments EG and GH are computed using the Law of Cosines.
When we get to points F, G, H, we find that there is some inconsistency with the locations that would be computed working clockwise using AB and angle BFA. This inconsistency shows up in an error in the 119° angle in the attached figure.
This means that segments on the back side of the route, along path EFGHA, will vary somewhat depending on how they're computed.
We assume that points B, C, F, G are collinear.
