Hi, I need help with this question I have. It's about basic trigonometric ratios, the question is in the attachment. I don't understand exactly what to do. Please help, thanks in advanced :)

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- Aug 13th 2012, 04:47 PMdan713Trigonometric Ratio
Hi, I need help with this question I have. It's about basic trigonometric ratios, the question is in the attachment. I don't understand exactly what to do. Please help, thanks in advanced :)

- Aug 14th 2012, 02:51 AMJohnDMalcolmRe: Trigonometric Ratio
The trig ratios can be remembered as SOH, CAH, TOA

$\displaystyle \sin{\theta} = \frac{\mbox{opposite}}{\mbox{hypoteneuse}} $

$\displaystyle \cos{\theta} = \frac{\mbox{adjacent}}{\mbox{hypoteneuse}} $

$\displaystyle \tan{\theta} = \frac{\mbox{opposite}}{\mbox{adjacent}} $

For the angle B, the adjacent arm of the triangle has a length 19 and the hypoteneuse has a length 27. These can be used directly to get the cosine ratio. To find the other two ratios you need the length of the arm opposite to B. Can you think of a formula that allows you to find this length? - Aug 14th 2012, 08:47 AMdan713Re: Trigonometric Ratio
Pythagorean theorem, thanks.