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Math Help - Statue of Liberty

  1. #1
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    Statue of Liberty

    A ship is just offshore of New York City. A sighting is taken of the statue of Liberty, which is about 305 feet tall. If the angle of elevation to the top of the statue is 20 degrees, how far is the ship from the statue?
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    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by magentarita View Post
    A ship is just offshore of New York City. A sighting is taken of the statue of Liberty, which is about 305 feet tall. If the angle of elevation to the top of the statue is 20 degrees, how far is the ship from the statue?
    Here
    You can form a triangle joining the ship(C) to the base of statue of liberty(B) and its top of statue(A) which is again joined to the ship(C)
     <br />
AB = \frac{BC}{tan(20)}<br />
  = 305 cot(20)<br />
----that's your answer.
    Where was the problem?
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    Do you.........

    Quote Originally Posted by ADARSH View Post
    Here
    You can form a triangle joining the ship(C) to the base of statue of liberty(B) and its top of statue(A) which is again joined to the ship(C)
     <br />
AB = \frac{BC}{tan(20)}<br />
= 305 cot(20)<br />
----that's your answer.
    Where was the problem?
    Do you mean your answer is already in lowest terms? There is nothing else to be done, right?
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  4. #4
    Like a stone-audioslave ADARSH's Avatar
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    Smile yes

    Quote Originally Posted by magentarita View Post
    Do you mean your answer is already in lowest terms? There is nothing else to be done, right?
    Yes , if you want to go ahead
    put
    the value cot(20)
    tan(20)=0.3640.......<br />
cot(20)=\frac{1}{0.3640}=2.7475
    put it in the answer and that's the of your answer!
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by ADARSH View Post
    Here
    You can form a triangle joining the ship(C) to the base of statue of liberty(B) and its top of statue(A) which is again joined to the ship(C)
     <br />
AB = \frac{BC}{tan(20)}<br />
= 305 cot(20)<br />
----that's your answer.
    Where was the problem?
    No need to introduce cotangent here. Most calculators don't have a COT key, so why use the reciprocal when good ole TAN will do nicely?

    \tan20^{\circ}=\frac{305}{x}

    x \tan20^{\circ}=305

    x=\frac{305}{\tan 20^{\circ}}

    x\approx 837.98 feet from the statue.
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  6. #6
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    ok but...

    Quote Originally Posted by ADARSH View Post
    Yes , if you want to go ahead
    put
    the value cot(20)
    tan(20)=0.3640.......<br />
cot(20)=\frac{1}{0.3640}=2.7475
    put it in the answer and that's the of your answer!
    My calculator does not have a cotagent key.

    Thanks
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  7. #7
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    thanks...........

    Quote Originally Posted by masters View Post
    No need to introduce cotangent here. Most calculators don't have a COT key, so why use the reciprocal when good ole TAN will do nicely?

    \tan20^{\circ}=\frac{305}{x}

    x \tan20^{\circ}=305

    x=\frac{305}{\tan 20^{\circ}}

    x\approx 837.98 feet from the statue.
    I thank you sincerely.
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