# question

• Jan 8th 2007, 10:16 AM
jimmybob
question
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 AM
ThePerfectHacker
Quote:

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.
• Jan 8th 2007, 10:24 AM
earboth
Quote:

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
• Jan 8th 2007, 10:27 AM
earboth
Quote:

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