Simpiifying trigonometric functions...urgent help needed

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Samara

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Oct 29, 2007
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I am having a really hard time with these problems...I think that I have got the first 3 correct (although I'm not completely positive)...I am stuck on the others...could someone please assist me through one of these problems?
11.jpg
 
Problems 17 and 18 are correct.

I remember problem 20 was solved for you - sometime ago.

Well -- you changed your problems now!!!
 
Samara said:
I am having a really hard time with these problems...I think that I have got the first 3 correct (although I'm not completely positive)...I am stuck on the others...could someone please assist me through one of these problems?
11.jpg
Click to expand...


(4) use tan t = (sin t )/(cos t)

(5) & (6) use \(\displaystyle cos^2x + sin^2x= 1\)

(7) use \(\displaystyle sec^2x = 1 + tan^2x\) and similar equation for (8)
 
Hello, Samara!

The first two problems are correct . . . Nice work!

If you solved the third one, I can't see your solution.

20) A flagpole is mounted on the front of a library's roof.

From a point 100 feet in front of the library, the angles of elevation
to the base of the flagpole and the top of the flagpole are 20[sup:efo9ncyi]o[/sup:efo9ncyi] and 39[sup:efo9ncyi]o[/sup:efo9ncyi]45', respectively.

Find the length of the flagpole.
Click to expand...

I'll make two diagrams for this one.

Code: 
    A *
      |
    f |
      |
    B *
      |  *
      |     *
    h |        *
      |           *
      |          20  *
    C * - - - - - - - - - * D
              100

The flagpole is: \(\displaystyle f = AB\)
The height of the library is: \(\displaystyle h = BC\)
The observer is at \(\displaystyle D:\;\angle BDC = 20^o\)

\(\displaystyle \text{We have: }\;\tan20^o \:=\:\frac{h}{100}\quad\Rightarrow\quad h \:=\:100\cdot\tan20^o\) .[1]


Code: 
    A *
      | *
    f |   *
      |     *
    B *       *
      |         *
      |           *
    h |             *
      |               *
      |          39.75  *
    C * - - - - - - - - - * D
              100

\(\displaystyle \angle ADC \,=\,39.75^o\)

\(\displaystyle \text{We have: }\;\tan39.75^o \:=\:\frac{f+h}{100}\quad\Rightarrow\quad f+h \:=\:100\cdot\tan39.75^o\quad\Rightarrow\quad f \:=\:100\cdot\tan39.75^o - h\)


Substitute [1]: .\(\displaystyle f \;=\;100\cdot\tan39.75^o - 100\cdot\tan20^o \;\approx\;46.77\) feet.

 
Hey I appreciate all of your help....but where are those problems coming from? I posted problem numbers 1-8 and I haven't changed my post since I posted it...(someone said that I posted that problem already, the flagpole...i don't know why that one is showing up...that was posted a long time ago)...sorry for any confusion...I don't know what's going on... :cry:
 
Samara said:
Hey I appreciate all of your help....but where are those problems coming from? I posted problem numbers 1-8 and I haven't changed my post since I posted it...(someone said that I posted that problem already, the flagpole...i don't know why that one is showing up...that was posted a long time ago)...sorry for any confusion...I don't know what's going on... :cry:
Click to expand...

Now, however, I have posted hints for 4-8. Did you get those?
 
Yes, thank you for you help! I was just wanted to apologize for any confusion and thank everyone who helped me. :D
 
Samara said:
I was just wanted to apologize for any confusion...
Click to expand...
No prob! :D

Note: I think the difficulty was a transient error caused when the site owner updated the forum script to a newer version. Somehow the wrong image was called...? I don't think you did anything "wrong"; it was probably just some "crossed wires"! :wink:

I hope you're having a great Thanksgiving! 8-)

Eliz.
 
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