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2. a schematic diagram of a pedestrian overpass is shown in the figure. If you walk on the overpass from one end to the other, how far have you walked? see attachment for details.
For this one , the awswer should be greater than 200 ft , yet my anwser is equal to 114.06 ft . Any help would be appreciated!
please don't use either law of sines or law of cosines to solve this problem ! Thanks
Let x be the first slaned distance (hypotenuse) travelled in 15 degree triangle. Let y be horizontal distanceQuote:
Originally Posted by vance
$\displaystyle \sin 15^{\circ} = \frac{18}{x}$
x = 69.546 ft
$\displaystyle \tan 15^{\circ} = \frac{18}{y}$
y = 67.177 ft
Let m be the second slaned distance (hypotenuse) travelled in 21 degree triangle. Let n be horizontal distance of triangle.
$\displaystyle \sin 21^{\circ} = \frac{18}{m}$
m = 50.228 ft
$\displaystyle \tan 21^{\circ} = \frac{18}{n}$
n = 46.892 ft
The middle horizontal distance travelled = 200 - 67.177 - 46.892 = 85.931 ft
Total distance travelled on bridge = 69.546 + 85.931 + 50.228 = 205.705 ft
