How to find the side of the square ,by using trigonometry

Question

There is a square with diagonal length of 'a", the question is to find the length of the sides. it can be found by Pythagorean theorem. but I tried to do it with trigonometry, considering the properties of a square but it doesn't resemble the first answer, so I was wondering what did I forget/mistake?

Answer 1

Hint:

A side is an orthogonal projection of the diagonal, onto a line making an angle of $\pi/4$ with the diagonal.

Answer 2

Let the side length of the square equal $b$, and the hypotenuse $a$. By Pythagoras' Theorem, we have

\begin{align} b^2+b^2&=a^2 \\ 2b^2&=a^2 \\ b^2&=\frac{a^2}{2}\\ b&=\sqrt{\frac{a^2}{2}}=\frac{\sqrt{a^2}}{\sqrt 2}=\frac{a}{\sqrt 2}=\frac{\sqrt 2}{2}a \end{align}

Alternatively, the angle between $a$ and $b$ is $45$ degrees. We know this because a square has $4$ right angles, which are bisected (cut in half) by the diagonal. Therefore,

\begin{align} \sin(45)&=\frac{b}{a}\\ \end{align}

Now find the value of $\sin(45)$, and rearrange the equation to find $b$. Hopefully this answer is familiar.

It seems that part of your confusion stems from the fact that you are unsure about rationalising the denominator:

$$ \frac{1}{\sqrt 2}=\frac{\sqrt 2}{2} $$

Consider the left-hand side of the above equality. Multiplying $\frac{1}{\sqrt 2}$ by $1$ does not change its value:

$$ \frac{1}{\sqrt 2}\times1=\frac{1}{\sqrt 2} $$

We also know that

$$ 1=\frac{\sqrt{2}}{\sqrt{2}} $$

Therefore,

$$ \frac{1}{\sqrt 2}\times1=\frac{1}{\sqrt 2}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt 2}{2} $$

Answer 3

The diagonal makes a $45^\circ=\frac{\pi}{4}$ angle with all vertices it touches with its endpoints.

If you look at trig identities here, you will see that $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\approx 0.707$ no matter which corner or which side of the diagonal you chose. The length of each side is $\frac{\sqrt{2}a}{2}$.