# finding sine, cosine, and tangent of theta

• Apr 8th 2011, 11:32 AM
sydewayzlocc
finding sine, cosine, and tangent of theta
I have a right triangle with x, y, and z is the hypotenuse. It has angles A and B.The given information is that z=17, A=theta, and B=theta. I used the pythagorean theorem to deduce that the sides are equal length at sqrt(289/2). I am to find sin of theta which is the opposite side over the hypotenuse, and is sqrt(289/2)/17 which reduces to 1/sqrt(2) i am unsure how it goes from sqrt(289/2)/17 to 1/sqrt(2) could someone please explain?
• Apr 8th 2011, 12:51 PM
masters
Quote:

Originally Posted by sydewayzlocc
I have a right triangle with x, y, and z is the hypotenuse. It has angles A and B.The given information is that z=17, A=theta, and B=theta. I used the pythagorean theorem to deduce that the sides are equal length at sqrt(289/2). I am to find sin of theta which is the opposite side over the hypotenuse, and is sqrt(289/2)/17 which reduces to 1/sqrt(2) i am unsure how it goes from sqrt(289/2)/17 to 1/sqrt(2) could someone please explain?

Hi sydewayzlocc,

I'd be happy to explain.

Two ways...

First, you have the legs at $\displaystyle \sqrt{\frac{289}{2}}$

That simplifies to $\displaystyle \frac{\sqrt{289}}{\sqrt{2}}=\frac{17}{\sqrt{2}}$

$\displaystyle \sin \theta = \dfrac{\frac{17}{\sqrt{2}}}{17}$

$\displaystyle \sin \theta = \dfrac{17}{\sqrt{2}} \times \dfrac{1}{17}=\dfrac{1}{\sqrt{2}}$

You would probably want to "rationalize" that answer: $\displaystyle \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}$

Or you could rationalize the lengths of the two legs first:

$\displaystyle \sqrt{\frac{289}{2}\times \frac{2}{2}} = \sqrt {\frac{17^2 \times 2}{4}}=\frac{17\sqrt{2}}{2}$

$\displaystyle \sin \theta = \dfrac{\frac{17\sqrt{2}}{2}}{17}$

$\displaystyle \sin \theta = \dfrac{\sqrt{2}}{2}$ which is the same as $\displaystyle \dfrac{1}{\sqrt{2}}$