Why Is Cosine Used to Calculate Angle Y in a Right Triangle?

[ai93]

Hi, this would be my first post of many in recent times as I have my maths exam soon! I am doing a lot of past papers and I need some help understand some questions.

In the triangle \(\displaystyle XYZ\) angle \(\displaystyle Z\) is a right angle. If \(\displaystyle XY\)= 15mm and \(\displaystyle YZ\)=8mm, calculate the angle \(\displaystyle Y\), giving your answer in degrees accurate to 1dp.

The solution was;
\(\displaystyle COSY=\frac{8}{15}\)
\(\displaystyle \therefore Y=COS^{-1}\frac{8}{15}\)
=\(\displaystyle Y=57.8\)

Can someone clarify why COS was used, and how to attempt similar questions like this?

Continuing from this question,
Calculate the length of the side \(\displaystyle XZ\)

I see pythag theorem was used as the solution was
\(\displaystyle XZ^{2}=\sqrt{15^{2}-8^{2}}\)
\(\displaystyle XZ\)= \(\displaystyle \sqrt{161} = 12.7\)

but why?

 

[MarkFL]

In the given triangle, we know it is a right triangle, and we know the hypotenuse is:

\(\displaystyle \overline{XY}=15\text{ mm}\)

And we also know the side adjacent to $\angle Y$ is:

\(\displaystyle \overline{YZ}=8\text{ mm}\)

Now, since the cosine function is defined as:

\(\displaystyle \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\)

we may then state:

\(\displaystyle \cos(Y)=\frac{8\text{ mm}}{15\text{ mm}}=\frac{8}{15}\)

And so we find:

\(\displaystyle Y=\arccos\left(\frac{8}{15}\right)\approx57.8^{\circ}\)

edit: To answer the added part, by Pythagoras, we know the sum of the squares of the legs is equal to the square of the hypotenuse, which allows us to write:

\(\displaystyle \overline{XZ}^2+8^2=15^2\)

\(\displaystyle \overline{XZ}=\sqrt{15^2-8^2}=\sqrt{161}\)

 

[ai93]

In the given triangle, we know it is a right triangle, and we know the hypotenuse is:

\(\displaystyle \overline{XY}=15\text{ mm}\)

And we also know the side adjacent to $\angle Y$ is:

\(\displaystyle \overline{YZ}=8\text{ mm}\)

Now, since the cosine function is defined as:

\(\displaystyle \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\)

we may then state:

\(\displaystyle \cos(Y)=\frac{8\text{ mm}}{15\text{ mm}}=\frac{8}{15}\)

And so we find:

\(\displaystyle Y=\arccos\left(\frac{8}{15}\right)\approx57.8^{\circ}\)

edit: To answer the added part, by Pythagoras, we know the sum of the squares of the legs is equal to the square of the hypotenuse, which allows us to write:

\(\displaystyle \overline{XZ}^2+8^2=15^2\)

\(\displaystyle \overline{XZ}=\sqrt{15^2-8^2}=\sqrt{161}\)

Thank you. Much clearer now. Since the question involves a right angle, we must use the Pythag Theorem. SOHCAHTOA helps a lot in solving for Y too! :D