pre-cal, formula and identity

[minito6]

6. The slope of a line is the tangent of the angle of inclination (the angle made with the positive x-axis). If the slope of one line is m1, and the slope of another line is m2, find a formula for tan(?2 ? ?1).

I honestly don't know where to start with this one, and am not exactly sure what the prof is looking for.

24. El Capitan is a large monolithic granite peak that rises straight up from the floor of Yosemite Valley in Yosemite National Park. It attracts rock climbers worldwide. At certain times the reflection of the peak can be seen in the Merced River that runs along the valley floor. Let H be the height of El Capitan, let h be the height of the sextant, let ? be the angle of elevation to the peak of El Capitan, and let ? be the angle of depression to the reflected peak in the river. Show that H = h [(1 + tan(?)cot(?))/(1 ? tan(?)cot(?))].

divide by h=H/h=1
cot a=(1/tan a)
so , [(1+tan B (1/tan a))/(1-tan B (1/tan a))]
tans cancel, so [(1+Ba)/(1-Ba)]

That's as far as I got, but I'm not sure if it's right, or I can do those steps.

14. A satellite S in circular (pretend) orbit around the Earth, is sighted by a tracking station T. The distance TS is determined by RADAR to be 1034 miles, and the angle of elevation above the horizon is 32.4?. How high is the satellite above the Earth at the time of the sighting if the radius of the Earth is 3964 miles?

I drew a picture, but can't figure out where to start.

 

[Deleted member 4993]

6. The slope of a line is the tangent of the angle of inclination (the angle
made with the positive x-axis). If the slope of one line is m1, and the
slope of another line is m2, find a formula for tan(?2 ? ?1).

Expand tan(?[sub:2ssnjrdv]2[/sub:2ssnjrdv] ? ?[sub:2ssnjrdv]1[/sub:2ssnjrdv])

How much is tan( ?[sub:2ssnjrdv]1[/sub:2ssnjrdv])?

How much is tan( ?[sub:2ssnjrdv]2[/sub:2ssnjrdv])?

 

[soroban]

Hello, minito6!

24. El Capitan is a large monolithic granite peak in Yosemite National Park.
At certain times the reflection of the peak can be seen in the Merced River.
Let \(\displaystyle H\) be the height of El Capitan.
Let \(\displaystyle h\) be the height of the sextant.
Let \(\displaystyle \beta\) be the angle of elevation to the peak of El Capitan.
Let \(\displaystyle \alpha\) be the angle of depression to the reflected peak in the river.

\(\displaystyle \text{Show that: } \;H \;= \; h\,\frac{\tan\alpha + \tan\beta}{\tan\alpha - \tan\beta}\) . . . . Rewritten for my convenience

                              C
                              *   -
                        *   * |   :
                  *       *   |   :
            * b         *     |   :
    A * - - - - - - - * - - - +F  :H
      | * a         *         |   :
     h|   *       *           |   :
      |   a *   * a           |   :
      * - - - * - - - - - - - *   -
      B       E               D

The transit is: \(\displaystyle AB = h\)
El Capitan is: \(\displaystyle CD = H\)

Draw \(\displaystyle AF \parallel BD\)
\(\displaystyle \alpha = \angle FAE\)
. . Note that: \(\displaystyle \angle AEB = \angle CED = \alpha\)
\(\displaystyle \beta = \angle CAF\)

\(\displaystyle \text{In right triangle }ABE\!:\;\tan\alpha = \frac{BE}{h}\quad\Rightarrow\quad BE \:=\:\frac{h}{\tan\alpha}\)

\(\displaystyle \text{In right triangle }CDE\!: \tan\alpha = \frac{ED}{H}\quad\Rightarrow\quad ED \:=\:\frac{H}{\tan\alpha}\)

\(\displaystyle \text{In right triangle }CFA\!:\;\tan\beta = \frac{CF}{AF}\quad\Rightarrow\quad AF \:=\:\frac{CF}{\tan\beta}\)
. . \(\displaystyle \text{Hence: }\;BD \;=\;\frac{H-h}{\tan \beta}\)

\(\displaystyle \text{Since }BD \;=\;BE + ED\text{, we have: }\;\frac{H-h}{\tan\beta} \:=\:\frac{h}{\tan\alpha} + \frac{H}{\tan\alpha} \;=\;\frac{H+h}{\tan\alpha}\)

. . \(\displaystyle (H-h)\tan\alpha \;=\;(H+h)\tan\beta \quad\Rightarrow\quad H\tan\alpha - h\tan\alpha \;=\;H\tan\beta + h\tan\beta\)

. . \(\displaystyle H\tan\alpha - H\tan\beta \;=\;h\tan\alpha + h\tan\beta \quad\Rightarrow\quad H(\tan\alpha - \tan\beta) \;=\;h(\tan\alpha + \tan\beta)\)

\(\displaystyle \text{Therefore: }\;\boxed{H \;=\;h\,\frac{\tan\alpha + \tan\beta}{\tan\alpha-\tan\beta}}\)

 

[minito6]

6. The slope of a line is the tangent of the angle of inclination (the angle
made with the positive x-axis). If the slope of one line is m1, and the
slope of another line is m2, find a formula for tan(?2 ? ?1).

Expand tan(?[sub:qk1r2of4]2[/sub:qk1r2of4] ? ?[sub:qk1r2of4]1[/sub:qk1r2of4])

How much is tan( ?[sub:qk1r2of4]1[/sub:qk1r2of4])?

How much is tan( ?[sub:qk1r2of4]2[/sub:qk1r2of4])?

The problem doesn't give a value for ? or m<sub>1 or m2.

So is that all I have to do is expand tan(?2 ? ?1)?</sub>

 

[stapel]

The problem doesn't give a value for ? or m1 or m2.

So is that all I have to do is expand tan(?2 ? ?1)?

Did the exercise ask you to "evaluate" the expression, but forgot to give you the values at which you're supposed to evaluate?

Or did the exercise ask you to find an expression for something, using the definitions and relationships provided in the exercise?

I know "soroban" merely handed you the solution for (24), but let's try to learn with (6) by doing some thinking and growing of our own!

So re-read the exercise; specifically, study the discussion of "the slope 'm[sub:16ef8brq]i[/sub:16ef8brq]' of a line is the tangent of the angle '@[sub:16ef8brq]i[/sub:16ef8brq]' of inclination of that line, so m[sub:16ef8brq]i[/sub:16ef8brq] = tan(@[sub:16ef8brq]i[/sub:16ef8brq])". Then follow the tutor's suggested steps, and see where the reasoning leads you. :wink:

Eliz.

P.S. For (14), please describe, in detail, the picture you drew, and provide whatever formulas or assumptions they gave you in class.

 

[minito6]

No, the problem just asks for this...

6. The slope of a line is the tangent of the angle of inclination (the angle
made with the positive x-axis). If the slope of one line is m1, and the
slope of another line is m2, find a formula for tan(?2 ? ?1).

That's it. So I'm assuming I need to find a formula that shows the lines m1 and m2 from their slopes, relating to tan(?2 ? ?1). Maybe in the form of mx+b? Would that be the right direction to take?

 

[stapel]

6. The slope of a line is the tangent of the angle of inclination (the angle made with the positive x-axis). If the slope of one line is m1, and the slope of another line is m2, find a formula for tan(?2 ? ?1).

That's it. So I'm assuming I need to find a formula that shows the lines m1 and m2 from their slopes...

They have given you the slope expressions, given you the angle expressions, and have told you that the slopes are the tangents. The tutor gave you the identity for expressing the required tangent in terms of the given slopes and angles. :!:

Where do you see line equations or line names in this exercise? Why are you trying to find slope values in terms of the lines? Why not use what's been provided, follow the instructions, and plug the given expressions into the given relationship? :shock:

I'm sorry, but I just don't understand what it is that you're trying to do. I mean, if you want to solve this exercise in some completely other manner than the straightforward method provided by the exercise and the tutor, that's fine, but it would really help if you explained what the other thing is that you're wanting to do. Thank you.

Eliz.

 

[Deleted member 4993]

6. The slope of a line is the tangent of the angle of inclination (the angle
made with the positive x-axis). If the slope of one line is m1, and the
slope of another line is m2, find a formula for tan(?2 ? ?1).

Expand tan(?[sub:3giaevy3]2[/sub:3giaevy3] ? ?[sub:3giaevy3]1[/sub:3giaevy3])................................(1)

How much is tan( ?[sub:3giaevy3]1[/sub:3giaevy3])? = m[sub:3giaevy3]1[/sub:3giaevy3]

How much is tan( ?[sub:3giaevy3]2[/sub:3giaevy3])?= m[sub:3giaevy3]2[/sub:3giaevy3]

Now substitute m[sub:3giaevy3]1[/sub:3giaevy3] and m[sub:3giaevy3]2[/sub:3giaevy3] for tan( ?[sub:3giaevy3]1[/sub:3giaevy3]) and tan( ?[sub:3giaevy3]1[/sub:3giaevy3]) respectively, in (1) and simplify.