Write the ratios for
sin
X
sin
X
and
cos
X
cos
X
.
Right triangle XYZ with right angle Z is shown. XY is labeled 12. XZ is labeled 5. YZ is labeled start root 119 end root.
nerd
There seem to be some uncertainties in the given question, but I will try to answer based on the information provided.
Assuming X is the angle opposite the side labeled 12, we can use the trigonometric ratios of sine and cosine to find the values of sin X and cos X.
sin X = opposite/hypotenuse = YZ/XZ = sqrt(119)/5
cos X = adjacent/hypotenuse = XY/XZ = 12/5
To write the ratios for sin X/sin X and cos X/cos X, we can simplify as follows:
sin X/sin X = 1
cos X/cos X = 1
Therefore, the ratios are both equal to 1, which means that sin X and cos X are equal to each other (assuming that the given angle X is acute).