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Mathematics 7 Online
OpenStudy (anonymous):

(Trigonometry) I'm not understanding the steps to solving this problem, may someone please help? P(-5,5) is a point on the terminal side of theta in standard position. What is the exact value of secant theta?

OpenStudy (anonymous):

|dw:1340061001411:dw| I got 7 using the pythagorean therom.

OpenStudy (anonymous):

Consider the definition of secent. Secent is inverse cosine, and cosine is \[\frac{\text{adjacent}}{\text{hypotenuse}}\] Which means that secent will be \[\frac{\text{hypotenuse}}{\text{adjacent}}\]

OpenStudy (anonymous):

Okay, so it'd be 7/5, yes?

OpenStudy (anonymous):

Good work =)

OpenStudy (mertsj):

Clearly the secant (reciprocal of the cosine) will be negative in quadrant 2. Calculate the hypotenuse using the fact that the hypotenuse of a 45-45-90 triangle if leg multiplied by sqrt2

OpenStudy (anonymous):

Oh, wait. One minor issue. -5 is the adjacent side.

OpenStudy (anonymous):

Right (: Because of the coordinates

OpenStudy (anonymous):

Yes. Well done.

OpenStudy (mertsj):

That cannot be correct. The hypotenuse is 5sqrt2 and so the secant is -5sqrt2/5or -sqrt2

OpenStudy (anonymous):

I know, it's just easier for me to start off with the number than it is with the square roots.

OpenStudy (anonymous):

Ah, yes. Mertsj is correct. You made some mistake in calculating the hypotenuse.

OpenStudy (mertsj):

Well, you can make up any answer you want but it will not be correct.

OpenStudy (anonymous):

Ah, alrighty. How do I work with the 5 square root 2?

OpenStudy (anonymous):

Just leave it as it is.

OpenStudy (anonymous):

still dd not get ? o.0

OpenStudy (mertsj):

\[\sec x=\frac{hypotenuse}{side adjacent}=\frac{5\sqrt{2}}{-5}=-\sqrt{2}\]

OpenStudy (anonymous):

Oh my gosh, THANK YOU :D That makes sense to me Lol.

OpenStudy (mertsj):

yw

OpenStudy (anonymous):

Nora, do you understand how to get the hypotenuse?

OpenStudy (anonymous):

That is the only thing I am still confused on :/

OpenStudy (mertsj):

Do you know the :Pythagorean Theorem?

OpenStudy (mertsj):

\[5^2+5^2=hypotenuse^2\]

OpenStudy (anonymous):

Yes, \[a ^{2} + b ^{2} = c ^{2}\]

OpenStudy (anonymous):

You were right that we need to use the pythagorean theorem. Set up an equation using a^2 + b^2 = c^2 and make sure that you use the hypotenuse for c.

OpenStudy (anonymous):

Okay, 5^2 + (-5)^2 = c^2 25 + 25 = c^2 50 = c^2

OpenStudy (anonymous):

Good. Keep going. Solve for c.

OpenStudy (anonymous):

\[\sqrt{50}\] = 7.07

OpenStudy (mertsj):

The problem asked for the EXACT value so you have to keep it in simplified radical form

OpenStudy (anonymous):

Well. Yes, sqrt(50) but this is where it's better not to plug it into your calculator and round it like you did.

OpenStudy (mertsj):

\[\sqrt{50}=\sqrt{25(2)}=5\sqrt{2}\]

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