Question

Look at the triangle.
What is the value of sin x°?

Answer 1

Look at the triangle.

Answer 2

do you know the mnemonic SOHCAHTOA?

SOH: Sine is Opposite / Hypotenuse
CAH: Cosine is Adjacent / Hypotenuse
TOA: Tangent is Opposite / Adjacent

You have a right triangle there, and you know the lengths of all of the sides. You should be able to use one of those 3 definitions to find an expression for the value of \(\sin x^\circ\)

Answer 3

Please explain that a bit, I haven't done this since elementary school.

Answer 4

Okay, we'll start with some basics. Do you know how to recognize a right triangle?

Answer 5

yes, 90 degree angle. has square instead of partial circle in angle.

Answer 6

Wait, I think I remember... let's see... to find sine of an angle, divide the length of the side opposite the angle by the hypotenuse. So that would be 12/13. Am I right?

Answer 7

@whpalmer4

Answer 8

Yes

Answer 9

Yes, that's correct, \[\sin x^\circ = \frac{12}{13}\]

If you can remember the mnemonic, then it is just a matter of identifying the correct sides to use. In this triangle, we also have

\[\cos x^\circ = \frac{5}{13}\]and \[\tan x^\circ = \frac{12}{5}\]

An interesting property of \(\sin\) and \(\cos\) is that \[\sin^2 x + \cos^2 x = 1\]and you can see that here:

\[(\frac{12}{13})^2 + (\frac{5}{13})^2 = 1\]\[\frac{144}{169} + \frac{25}{169} =1\]\[\frac{144+25}{169}=1\]\[\frac{169}{169} = 1\]\[1=1\checkmark\]

Sorry I wasn't able to respond earlier!