2019-04-09
Apr 10 In-Class Exercise .
Post your solutions to the Apr 10 In-Class Exercise to this thread.
Best,
Chris
(Edited: 2019-04-09)Post your solutions to the Apr 10 In-Class Exercise to this thread.
Best,
Chris
2019-04-10
The height of the E would be 20tan(1/6) = 0.05 feet.
(Edited: 2019-04-10)The height of the E would be 20tan(1/6) = 0.05 feet.
We know that tan(theta)= e/d
e = tan(theta)*d = 0.05
We know that tan(theta)= e/d
e = tan(theta)*d = 0.05
tan(1/6 degrees)*20 = 0.05817780582 feet
tan(1/6 degrees)*20 = 0.05817780582 feet
height will be 20ft * tan(1/6 degrees) which equals 0.058 ft
height will be 20ft * tan(1/6 degrees) which equals 0.058 ft
20 * tan((1/6)/180 * pi) = 0.05817780569
@BT@20 * tan((1/6)/180 * pi) = 0.05817780569@BT@
s = d*tan(theta)
s = 20*tan(1/6 degrees) = 0.06 ft
(Edited: 2019-04-10)s = d*tan(theta)
s = 20*tan(1/6 degrees) = 0.06 ft
<nowiki> s = 20ft * tan(1/6 degrees) s = 20 ft * 0.00291 s = 0.058 </nowiki>
(Edited: 2019-04-10)<nowiki>
s = 20ft * tan(1/6 degrees)
s = 20 ft * 0.00291
s = 0.058
</nowiki>
1/6 of a degree = 0.00290888208665 radians s = 20 * tan (0.00290888208665) s = 0.058
1/6 of a degree = 0.00290888208665 radians
s = 20 * tan (0.00290888208665)
s = 0.058
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