1. ## Trig Question

To measure the Eiffel Tower in Paris,
a person stands away from the base and measures the angle of elevation to the top of the tower to be 60◦. Moving 210 feet closer, the angle of elevation to the top of the tower is 70◦. How tall is the Eiffel Tower?

why is it wrong to do cos(60) = (210/x) getting x = 420
& then doing cos(20) = (x/420) getting that the Eiffel tower = 395 ft. ???

Any help would be really appreciated! thank you.

2. the base where 210 is
is not the base of right triangle

i would use cot to solve this... where the bases are x and (210 + x)

$\displaystyle cot60\,^{\circ} = \frac{x+210}{h}$

$\displaystyle cot70\,^{\circ} = \frac{h}{x}$

$\displaystyle cot60\,^{\circ}-cot70\,^{\circ} = \frac{x + 210}{h}-\frac{x}{h} = \frac{210}{h}$

$\displaystyle h = \frac{210}{cot60\,^{\circ}- cot70\,^{\circ}} \approx 984ft$

Wikipedia says the Eiffel Tower is 1063ft

3. Originally Posted by VkL
To measure the Eiffel Tower in Paris,
a person stands away from the base and measures the angle of elevation to the top of the tower to be 60◦. Moving 210 feet closer, the angle of elevation to the top of the tower is 70◦. How tall is the Eiffel Tower?

why is it wrong to do cos(60) = (210/x) getting x = 420
One thing wrong with it is that there is no "x" in your picture so we have no idea what "x" means. More important is the fact that 210 is not a side of a right triangle. cos(60) should be "near side over hypotenuse" so I would guess you mean x to be the length of the hypotenuse (the straight line distance from the 60 degree angle to the top of the eiffel tower?) but "210" is NOT the length of a side of a right triangle.

& then doing cos(20) = (x/420) getting that the Eiffel tower = 395 ft. ???

Any help would be really appreciated! thank you.

4. Originally Posted by VkL
...why is it wrong to do cos(60) = (210/x) getting x = 420
& then doing cos(20) = (x/420) getting that the Eiffel tower = 395 ft. ???
It is wrong ONLY because it doesn't help in answering the question being asked.
Your value of 420 is the distance from the 60deg observation point toward the top of the tower.
I do not understand how you are approaching the problem.
Could you draw out what you "expected" your values to represent?

I would do it this way:
(see attached sketch)

$\displaystyle \text{Tower}\times \tan(30deg) = d + 210$
$\displaystyle \text{Tower}\times \tan(20deg) = d$

subtract 2nd equation from the first equation:

$\displaystyle \text{Tower}\times \tan(30deg) - \text{Tower}\times \tan(20deg) = 210$

&
$\displaystyle \text{Tower} (\, \tan(30deg) - \tan(20deg) \, ) = 210$

&
$\displaystyle \text{Tower} = \dfrac{210}{\tan(30deg) - \tan(20deg) }$ = 984.2

.