## Hard problem with right triangles

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
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### Hard problem with right triangles

I need help with number 1 on question 1 and number 1 on question 2

http://dl.dropbox.com/u/4592031/trigono ... ject_2.pdf

stapel_eliz
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dade488 wrote:I need help with number 1 on question 1 and number 1 on question 2

http://dl.dropbox.com/u/4592031/trigono ... ject_2.pdf

What have you tried? Where are you stuck?

Please be complete, so we can "see" where you are needing help. Thank you!

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### Re: Hard problem with right triangles

in problem number 1 i'm stuck with the first question. I don't understand how I can show that the height of the mountain is given by that specific formula. why am I using tangents of alpha and beta? I got number 2 and 3 for the first problem.

the second problem I can't figure out number 1 and I need that in order to figure out all the others (2 through 6). I can't find out tangent of theta because it's not a right triangle. I need to figure out how to get that final formula so I can proceed with the rest.
Makes sense?

stapel_eliz
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dade488 wrote:the second problem I can't figure out number 1....

You know that the tangent on the left-hand side of the identity they've given you is 13/x. You know that the tangent of alpha is 4/x.

Plug those in, solve for the tangent of theta, and then take the inverse tangent. You'll then have theta as a function of x.

stapel_eliz
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dade488 wrote:in problem number 1 i'm stuck with the first question. I don't understand how I can show that the height of the mountain is given by that specific formula. why am I using tangents of alpha and beta?

Because $\alpha$ and $\beta$ are the angles they've given you.

Okay, okay; I'll behave....

Label the picture with "m" being the whole height of the mountain, "s" being the height of the sextant, and "m - s" being the height of the mountain less the height of the sextant. Label the part of the lower horizontal line between the base of the sextant and where the angle line hits the horizontal as "x"; label the other half as "y". Then you have:

. . . . .$\frac{m}{y}\, =\, \tan(\alpha)$

. . . . .$\frac{s}{x}\, =\, \tan(\alpha)$

. . . . .$\frac{m\, -\, s}{x\, +\, y}\, =\, \tan(\beta)$

See if you can use the first two to substitute for the variables you don't want in the third, and then solve for "m".

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### Re:

stapel_eliz wrote:
dade488 wrote:in problem number 1 i'm stuck with the first question. I don't understand how I can show that the height of the mountain is given by that specific formula. why am I using tangents of alpha and beta?

Because $\alpha$ and $\beta$ are the angles they've given you.

Okay, okay; I'll behave....

Label the picture with "m" being the whole height of the mountain, "s" being the height of the sextant, and "m - s" being the height of the mountain less the height of the sextant. Label the part of the lower horizontal line between the base of the sextant and where the angle line hits the horizontal as "x"; label the other half as "y". Then you have:

. . . . .$\frac{m}{y}\, =\, \tan(\alpha)$

. . . . .$\frac{s}{x}\, =\, \tan(\alpha)$

. . . . .$\frac{m\, -\, s}{x\, +\, y}\, =\, \tan(\beta)$

See if you can use the first two to substitute for the variables you don't want in the third, and then solve for "m".

ok, that's as far as I got.
when I go to substitute though I get confused again...I feel like I can't plug m/y or s/x anywhere in the second...am I wrong?

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### Re:

stapel_eliz wrote:
dade488 wrote:the second problem I can't figure out number 1....

You know that the tangent on the left-hand side of the identity they've given you is 13/x. You know that the tangent of alpha is 4/x.

Plug those in, solve for the tangent of theta, and then take the inverse tangent. You'll then have theta as a function of x.

It would be 9/x because 9 is the length of the whole side. 4 is the length of the shorter side.
so this is what I am left with:

9/x= ((4/x)+tantheta)/(1-(4tantetha/x))
Is this right?

stapel_eliz
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dade488 wrote:ok, that's as far as I got.

Next time, it might be helpful to show that when you're asked "What have you tried? Where are you stuck?"

dade488 wrote:when I go to substitute though I get confused again...I feel like I can't plug m/y or s/x anywhere in the second...am I wrong?

Well, you could try using the suggestion provided earlier: Solve the first two equations for the variables you don't want, and then use these in the third equation.

dade488 wrote:so this is what I am left with:

9/x= ((4/x)+tantheta)/(1-(4tantetha/x))
Is this right?

As far as it goes. Now solve for $\tan(\theta)$, and take the inverse tangent.

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### Re: Re:

stapel_eliz wrote:
dade488 wrote:the second problem I can't figure out number 1....

You know that the tangent on the left-hand side of the identity they've given you is 13/x. You know that the tangent of alpha is 4/x.

Plug those in, solve for the tangent of theta, and then take the inverse tangent. You'll then have theta as a function of x.

It would be 9/x because 9 is the length of the whole side. 4 is the length of the shorter side.
so this is what I am left with:

9/x= ((4/x)+tantheta)/(1-(4tantetha/x))
Is this right?

If I keep going I get this:

tantheta= 5/(x+(36/x))...
??

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### Re: Re:

stapel_eliz wrote:Okay, okay; I'll behave....

Label the picture with "m" being the whole height of the mountain, "s" being the height of the sextant, and "m - s" being the height of the mountain less the height of the sextant. Label the part of the lower horizontal line between the base of the sextant and where the angle line hits the horizontal as "x"; label the other half as "y". Then you have:

. . . . .$\frac{m}{y}\, =\, \tan(\alpha)$

. . . . .$\frac{s}{x}\, =\, \tan(\alpha)$

. . . . .$\frac{m\, -\, s}{x\, +\, y}\, =\, \tan(\beta)$

See if you can use the first two to substitute for the variables you don't want in the third, and then solve for "m".

So I get :
tanbeta=(m-s)/((s+m)/tanalpha)

m=s+((tanbeta*s)/tanalpha)+((tanbeta*m)/tanalpha)