In a right angled triangle ABC Sin A = 5/13

calculate without finding the angel A the following

a) Cos A

b) Tan A

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- Jan 8th 2007, 10:16 AMjimmybobquestion
In a right angled triangle ABC Sin A = 5/13

calculate without finding the angel A the following

a) Cos A

b) Tan A - Jan 8th 2007, 10:23 AMThePerfectHacker
That means the opposite side is 5 and hypotenuse is 13 because sine is opposite over hypotenuse. That means by Pythagorean theorem the adjacent side is 12. Thus, cos A =12/13 because it is adjacent over opposite. And tangent is 5/12 because tangent is opposite over adjacent.

- Jan 8th 2007, 10:24 AMearboth
Hello,

use the formula:

$\displaystyle \sin^2(x)+\cos^2(x)=1$. Thus

$\displaystyle \left(\frac{5}{13} \right)^2+\cos^2(x)=1 \Longrightarrow \cos(A) = \frac{12}{13}$

to b)

$\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$. Thus

$\displaystyle \tan(A)=\frac{5}{12}$

EB - Jan 8th 2007, 10:27 AMearboth