# Statue of Liberty

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• Nov 13th 2008, 02:52 AM
magentarita
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?
• Nov 13th 2008, 05:06 AM
ADARSH
Quote:

Originally Posted by magentarita
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)
$\displaystyle AB = \frac{BC}{tan(20)} = 305 cot(20)$----that's your answer.
Where was the problem?(Nerd)
• Nov 13th 2008, 06:40 AM
magentarita
Do you.........
Quote:

Originally Posted by ADARSH
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)
$\displaystyle AB = \frac{BC}{tan(20)} = 305 cot(20)$----that's your answer.
Where was the problem?(Nerd)

Do you mean your answer is already in lowest terms? There is nothing else to be done, right?
• Nov 13th 2008, 10:25 PM
ADARSH
yes
Quote:

Originally Posted by magentarita
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)
$\displaystyle tan(20)=0.3640....... cot(20)=\frac{1}{0.3640}=2.7475$
put it in the answer and that's the (Makeup) of your answer!
• Nov 14th 2008, 09:47 AM
masters
Quote:

Originally Posted by ADARSH
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)
$\displaystyle AB = \frac{BC}{tan(20)} = 305 cot(20)$----that's your answer.
Where was the problem?(Nerd)

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?

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

$\displaystyle x \tan20^{\circ}=305$

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

$\displaystyle x\approx 837.98$ feet from the statue.
• Nov 14th 2008, 01:02 PM
magentarita
ok but...
Quote:

Originally Posted by ADARSH
:) Yes , if you want to go ahead
put
the value cot(20)
$\displaystyle tan(20)=0.3640....... cot(20)=\frac{1}{0.3640}=2.7475$
put it in the answer and that's the (Makeup) of your answer!

My calculator does not have a cotagent key.

Thanks
• Nov 14th 2008, 01:03 PM
magentarita
thanks...........
Quote:

Originally Posted by masters
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?

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

$\displaystyle x \tan20^{\circ}=305$

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

$\displaystyle x\approx 837.98$ feet from the statue.

I thank you sincerely.