MHB How far above the ground is the tack?

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Devon's bike has 27-inch diameter wheels, and after rolling 100 feet, he stops with a tack picked up by the front wheel. The circumference of the tire is approximately 84.82 inches, leading to a total distance traveled of 1200 inches, resulting in about 14.147 revolutions. A miscalculation initially led to an incorrect angle and height above the ground for the tack. The correct calculations reveal that the tack is approximately 5.368 inches above the ground after accounting for the wheel's rotation.
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Devon's bike has wheels that are 27 inches in diameter. After the front wheel picks up a tack, Devon rolls another 100 feet and stops. How far above the ground is the tack?

I did all my work carefully but the answer seems wierd.

circumference of tire :84.78 inches
total distance traveled is 1200 inches

1200 /84.78 = 14.154

84.78 : 0.154 = 360: X

X = 0.653

13.5cos0.653 = 13.4991

13.5 - 13.4991 = 0.000876

answer : 0.000876
 
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First, it would help to use more digits of pi. Then 1200/(27$\pi$)=14.147 up to 3 decimal digits.

veronica1999 said:
84.78 : 0.154 = 360: X
This should be X / 360 = 0.147 (this is a fraction of the full circle). Then X = 52.958 degrees and the final answer is 5.368 inches.
 
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Thank you! :D
Now I see my mistake.

Instead of 84.78 : 0.154 = 360 : X
the equation should have been 84.78 :13.056 = 360 :X
X = 55.4

1200/27pi = 14.147

84.78 : 12.46 = 360 :X
x= 52.958
 
Hello, veronica1999!

Devon's bike has wheels that are 27 inches in diameter.
After the front wheel picks up a tack, Devon rolls another 100 feet and stops.
How far above the ground is the tack?

I did all my work carefully but the answer seems weird.

Circumference of tire: 84.78 inches . This is wrong.
Total distance traveled is 1200 inches

1200 / 84.78 = 14.154

84.78 : 0.154 = 360 : X . This is wrong, too.

X = 0.653 . That is a VERY small angle, isn't it?

13.5cos0.653 = 13.4991

13.5 - 13.4991 = 0.000876

Answer: 0.000876
$\text{Circumference: }\:27\pi \:\approx\:84.82 $

$ \text{Then: }\:\dfrac{1200}{84.82} \:=\:14.1476\text{ revolutions} $

$\text{The wheel makes 14 revolutions and a fraction of a revolution.}$

$\text{That fraction is: }\:0.1476 \times 360^o \:=\:53.136^o $

$\text{Got it?} $
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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