Precalculus problem- sine, cosine, and inversecant

[AndyXUe]

Given that sin(x) = O/P and cos(x) = H/P, find: arcsec(-O/H)

Can anybody explain the answer to this and how they got this answer? This was a question on one of my quizzes and I was unable to solve this. Thanks for the help!

 

[tkhunny]

Did you draw a right triangle and label the parts?

1) Draw a right triangle.
2) Label an acute angle 'x'.
3) Given sin(x) = O/P, label the opposite legt "O" and the Hypotenuse "P"
4) Given cos(x) = H/P. lebel the adjacent leg "H". The hypotenuse is labeled already.

5) sec(x) = 1/cos(x) = P/H

That was the art project. The mathemaitcs is now to appear. Let's see what you get.

 

[AndyXUe]

sorry i meant find: arcsec(-O/H), not inverse, sorry for the confusion. i am stumped as to what to do to find arcsec(-O/H)

 

[tkhunny]

There was no confusion. Your inverse secant, or arcsec is the next step of this solution.

It remain that cosine is given and secant is its reciprocal.

 

[tkhunny]

Please don't double post.

If cos(x) = H/P. then sec(x) = P/H, then arcsec(P/H) = x

If sin(x) = O/P, then cos(90º - x) = O/P. then sec(90º - x) = P/O, then arcsec(P/O) = ????

Look at all the pieces. Fidn the ones that solve your problem.

 

[AndyXUe]

I got the answer to be x = 90 degrees because sec(90) = undefined and when you draw it out you can see the relation. Is this answer right? Thanks for the help!

 

[tkhunny]

90º is no good. You do not have any specific measurements, excepting the right angle.