Right Triangle Trig

[azure perch]

How would I go abouts even starting this?

This is what I think the graph should look like.

I assumed that I would use tan(9)=O/154, but that is definietly not right.

Oh, wait

[balmy ruin]

That's correct actually

[azure perch]

I figured it out, but why do I need to use the 81 degrees angle instead of 9?

[balmy ruin]

You just have to convert in radians

[azure perch]

154tan(81)=O gave me the correct answer

[balmy ruin]

You can either use the 81° angle or the 9° angle here

$\tan(9^\circ) = x/154$ and $\tan(81^\circ) = 154/x$

[tranquil viper]

Twenty

[balmy ruin]

This should not be correct though because tan is opposite over adjacent so for 81° the opposite is 154ft

[azure perch]

Ahh, got it

I'm a bit confused now with this type of logic

So let's say I have this question, c = sqrt(164)

Would the tan be 10/8 or 8/10?

It depends how you draw the graph, no?

Oh, no it doesn't. It would always be 10/8, right?

a is always the bottom line

Thank goodness, I really understand right triangle trig now. Thank you so much!

.close