SOH CAH TOA!!!!!! I will never forget that

I agree that the question could be more explicit on that point, but I think that the most reasonable interpretation is that it seeks the distance from the observer to the boat.

This is neither here nor there, but that point reminds me of a really good Yogi Berra quote. Someone asked Yogi what time it was. He said, "You mean now?"

OK math geeks, up next is a detailed discussion of Newton's method for solving complex polynomials and numerous linear algebra problems of about 20 variables.

Trust me, you're going to need this in life.


If you're a rocket scientist or economist.

My high water mark in both math and computer programming was a Paschal program I wrote to perform Newton's method on complex polynomials up to x to the power of 10. It was fun.

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Te futueo et caballum tuum

1986 300SDL, 362K
1984 300D, 138K

prime us first.

I get the math... but the concept and how it came to be still shakes me a little.

So I would take 300/(.2756)(15.9)

That actually was the original question.

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86 300SDL. 250,xxx on #14 Head. One eye always on temp gauge.. Cruising towards 300K

I just plugged that into a calculator and did not get that answer. I understand the concept but can not figure out how to do it long hand.

If I did not score high enough on the math portion I can retake it but I will need to know the steps of the calculation..

Algbra not an issue trig for me always an issue.

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86 300SDL. 250,xxx on #14 Head. One eye always on temp gauge.. Cruising towards 300K

you are using the sine function as a value - incorrect.

you 'take' the sine 'of' some angle or radian measure.---this gives you a numeric value.

Now you can see why i did it the other way . How do you take the sine? I am trying to get my head around what should be a simple calculation and am not getting it.

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86 300SDL. 250,xxx on #14 Head. One eye always on temp gauge.. Cruising towards 300K

Never mind the light bulb just came on. I would use the .2756 as the denominator under 300 to solve for x.

Which is the S=O
H

Correct? I hope.......

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86 300SDL. 250,xxx on #14 Head. One eye always on temp gauge.. Cruising towards 300K

so use the chart to get the values for sin of some angle. In this case 16* is your closest approximation on a timed exam with multiple choice.

yeppers.

.2756 from the chart is the value for sine of 16*

So another question that sticks in my mind uses the angle of the sun and the length of a shadow cast from a flag pole. seeing that I have the Hyp angle and length ,I would use the cos to find the short leg?

which in this case would be the height of the flagpole?

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86 300SDL. 250,xxx on #14 Head. One eye always on temp gauge.. Cruising towards 300K

in order to find the value of a trig function by hand, you would need to expand a Taylor series. No way thats happening on a timed exam with any accuracy.

you either have a slide rule, the tables or remember a bunch of values.

and you convert degrees to radians, when needed.

given: hyp = h ; theta= b

sin(b)=Opposite/h ==> sin(b) * h =opposite

I am assuming theta is the angle you're talking about. The sun is at the top right. "the short leg" is 'opposite'.

if the short leg is "adjacent"

cos(b) = adjacent/h ==> cos(b) * h = adjacent

Thank you all for the help.

Jt20 that picture will be filed. That is the piece of info I needed. I should not a have an issue if I need to take this thing over.

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86 300SDL. 250,xxx on #14 Head. One eye always on temp gauge.. Cruising towards 300K