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KengaRu [80]
3 years ago
12

An arch for a bridge over a highway is in the form of a semiellipse. The top of the arch is 35 feet above ground (the major axis

). What should the span of the bridge be (the length of its minor axis) if the height 25 feet from the center is to be 10 feet above ground?
Mathematics
1 answer:
jeka943 years ago
8 0

For given problem:

Put midpoint of ellipse, (0,0) at epicenter of bridge at ground level.

Specified length of vertical major axis = 70=2a

a=35

a^2=1225

 

Equation of ellipse:

 

x^2/b^2+y^2=1

 

plug in coordinates of given point on ellipse(25, 10)

 

25^2/b^2 + 10^2/a^2 = 1

 

625/b^2 + 100/1225=1

 

625/b^2 = 1 - 100/ 1225 = .918

 

b^2 = 625/.918 ≈ 681

 

b ≈ 26.09

 

length of minor axis = 2b = 2(26.09) ≈ 52.16 ft

 

Span of bridge ≈ 52.16 ft

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Hypotenuse =\sqrt{82}

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Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

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\csc^2(\theta) = (\csc(\theta))^2

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\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

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