(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?
|dw:1340061001411:dw| I got 7 using the pythagorean therom.
Consider the definition of secent. Secent is inverse cosine, and cosine is \[\frac{\text{adjacent}}{\text{hypotenuse}}\] Which means that secent will be \[\frac{\text{hypotenuse}}{\text{adjacent}}\]
Okay, so it'd be 7/5, yes?
Good work =)
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
Oh, wait. One minor issue. -5 is the adjacent side.
Right (: Because of the coordinates
Yes. Well done.
That cannot be correct. The hypotenuse is 5sqrt2 and so the secant is -5sqrt2/5or -sqrt2
I know, it's just easier for me to start off with the number than it is with the square roots.
Ah, yes. Mertsj is correct. You made some mistake in calculating the hypotenuse.
Well, you can make up any answer you want but it will not be correct.
Ah, alrighty. How do I work with the 5 square root 2?
Just leave it as it is.
still dd not get ? o.0
\[\sec x=\frac{hypotenuse}{side adjacent}=\frac{5\sqrt{2}}{-5}=-\sqrt{2}\]
Oh my gosh, THANK YOU :D That makes sense to me Lol.
yw
Nora, do you understand how to get the hypotenuse?
That is the only thing I am still confused on :/
Do you know the :Pythagorean Theorem?
\[5^2+5^2=hypotenuse^2\]
Yes, \[a ^{2} + b ^{2} = c ^{2}\]
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.
Okay, 5^2 + (-5)^2 = c^2 25 + 25 = c^2 50 = c^2
Good. Keep going. Solve for c.
\[\sqrt{50}\] = 7.07
The problem asked for the EXACT value so you have to keep it in simplified radical form
Well. Yes, sqrt(50) but this is where it's better not to plug it into your calculator and round it like you did.
\[\sqrt{50}=\sqrt{25(2)}=5\sqrt{2}\]
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