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## Homework Statement

Not long ago I had a similar problem, which I was able to solve after reading this thread, but for this question I'm stuck and I could use a small hint.

## Homework Equations

## The Attempt at a Solution

Thanks

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- Thread starter jkristia
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Not long ago I had a similar problem, which I was able to solve after reading this thread, but for this question I'm stuck and I could use a small hint.

Thanks

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- #3

SammyS

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Label the triangle differently. Label it so that the tangent of α is v .## Homework Statement

Not long ago I had a similar problem, which I was able to solve after reading this thread, but for this question I'm stuck and I could use a small hint.

## Homework Equations

## The Attempt at a Solution

Thanks

[itex]\displaystyle \tan(\alpha)=\frac{v}{1}\,.[/itex]

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>>Label the triangle differently. Label it so that the tangent of α is v .

hmm, I think that is what I have, tan (α) = v = sqrt..., I'm still missing something.

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>>Label the triangle differently. Label it so that the tangent of α is v .

hmm, I think that is what I have, tan (α) = v = sqrt..., I'm still missing something.

Okay, so you have [itex]tan(\alpha) = v[/itex]. What is [itex]arctan(v)[/itex]? Then, working from the inside out, what is [itex]sin(arctan(v))[/itex]?

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Look in your original picture. v is equal to four quantities, which one would give you a nice value for [itex]arctan(v)[/itex]?

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ah - I think I got it, will try later. I need to find the inverse of v = sqrt()/x.

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What is [itex]arctan(tan(x))[/itex] equal to?

Can you see which value to plug in now?

Then you need to calculate sine of that number you get, which is also listed in the picture as a number not involving trig functions or trig inverses.

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This is how I solved it. It is exactly the same steps as on the link. I read the explanation, closed the browser and solved it myself by first drawing a new trigangle with the 'correct' labels, and then it pretty much fell in place.

I think the trick is to label the triangle with the sides as v/1 = v for the inner trig function, and that is where I made the mistake.... I think

Again, thank you for your help.

Edit: I just noticed that the tanθ does not provide any additional information.

I think the trick is to label the triangle with the sides as v/1 = v for the inner trig function, and that is where I made the mistake.... I think

Again, thank you for your help.

Edit: I just noticed that the tanθ does not provide any additional information.

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