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.

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.

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.:(

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 :rolleyes:. 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.

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

Which is the S=O
H

Correct? I hope.......

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?

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.

http://www.foundation-flash.com/tuto...triglabels.png

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.