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**jimmybob** · January 8th 2007, 10:16 AM

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

calculate without finding the angel A the following

a) Cos A

b) Tan A

**ThePerfectHacker** · January 8th 2007, 10:23 AM

Originally Posted by jimmybob

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

calculate without finding the angel A the following

a) Cos A

b) Tan A

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.

**earboth** · January 8th 2007, 10:24 AM

Originally Posted by jimmybob

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

calculate without finding the angel A the following

a) Cos A

b) Tan A

Hello,

use the formula:

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

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

to b)

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

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

EB

**earboth** · January 8th 2007, 10:27 AM

Originally Posted by ThePerfectHacker

That means the opposite side is 5 and hypotenuse is 13 ....

Hello,

only to be very exact the opposite side is a multiple of 5 and the hypotenuse is the same multiple of 13....

EB

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