# Trigonometry!!

• Jan 13th 2009, 10:06 AM
Zakir
Trigonometry!!
I have had some question on trig and i could easily find the lines like e.g. whats the size of A-B
However when im doing this i either have to do SIN the angle ,TAN thhe angle,COS the angle.
But what happens if there is no angle.
This is what i do..
Firstly lets say
the angle is called "alpha" and we are using tangent aand opposite,
so we use TAN
next i did is
TAN "alpha" = lets say the sides are 6 which is the opposite and 3 the tangent
all i did was TAN "alpha" = 6/3
Now i did this on my calculator.
INVERSE TAN 3
iam shure i did everythign correct apart from last part so can someone explain please.
Thank you.
• Jan 13th 2009, 10:15 AM
Trigonometry
Hello Zakir
Quote:

Originally Posted by Zakir
I have had some question on trig and i could easily find the lines like e.g. whats the size of A-B
However when im doing this i either have to do SIN the angle ,TAN thhe angle,COS the angle.
But what happens if there is no angle.
This is what i do..
Firstly lets say
the angle is called "alpha" and we are using tangent aand opposite,
so we use TAN
next i did is
TAN "alpha" = lets say the sides are 6 which is the opposite and 3 the tangent
all i did was TAN "alpha" = 6/3
Now i did this on my calculator.
INVERSE TAN 3
iam shure i did everythign correct apart from last part so can someone explain please.
Thank you.

If $\displaystyle \tan \alpha = 6/3$, then you will need INVERSE TAN 2, because 6/3 = 2.

This gives $\displaystyle \alpha = 63.4^o$

Is that what you needed to know?

• Jan 13th 2009, 10:17 AM
masters
Quote:

Originally Posted by Zakir
I have had some question on trig and i could easily find the lines like e.g. whats the size of A-B
However when im doing this i either have to do SIN the angle ,TAN thhe angle,COS the angle.
But what happens if there is no angle.
This is what i do..
Firstly lets say
the angle is called "alpha" and we are using tangent aand opposite,
so we use TAN
next i did is
TAN "alpha" = lets say the sides are 6 which is the opposite and 3 the tangent
all i did was TAN "alpha" = 6/3
Now i did this on my calculator.
INVERSE TAN 3
iam shure i did everythign correct apart from last part so can someone explain please.
Thank you.

Hello Zakir,

I'm having a little trouble following what you're asking, but suffice it to say that the tangent function is the ratio of the oppsite side to the adjacent side of a right triangle.

$\displaystyle \tan \alpha = \frac{6}{3}$

$\displaystyle \arctan 2=\alpha$