MHB Acute angle of right triangles

[mathmari]

Hey!

We have a rectangle inside a semicircle with radius $1$ :

View attachment 9703

From the midpoint of the one side we draw a line to the opposite vertices and one line to the opposite edge.

View attachment 9704

Are the acute angles of the right triangles all equal to $45^{\circ}$ ? (Wondering)

All four triangles are similar, aren't they? We have that the hypotenuse of each right triangle is equal to $1$, since it is equal to the radius of the circle.
I am stuck right now about the angles. (Wondering)

 

[HallsofIvy]

A little thought should show you those statements are NOT true! There exist many different such rectangles, with many different such angles, depending on the height of the rectangle.

 

[mathmari]

A little thought should show you those statements are NOT true! There exist many different such rectangles, with many different such angles, depending on the height of the rectangle.

In this case the resulting smaller rectangles look like squares and that's why maybe I got confused. (Doh)

So when we know that the area of the big rectangle is $1$ and we want to calculate the length of the sides, it is not a good idea to use trigonometry, right? (Wondering)

It is better to do the following:

View attachment 9705

Let $x$ be the height and $w$ the width. Since $M$ is the midpoint we get that $w=2y$.
At the right triangle we can Pythagoras' Theorem and we get that $y=\sqrt{1-x^2}$.
The area of the big rectangle is $1$ so we get that $x\cdot w=1 \Rightarrow x\cdot 2\sqrt{1-x^2}=1$ and from that eauation we can calculate $x$. Btw we would get the same result if we would consider the acute angles to be $45^{\circ}$, so in this case they are indeed like that.

 

[I like Serena]

So when we know that the area of the big rectangle is $1$ and we want to calculate the length of the sides, it is not a good idea to use trigonometry, right?

Hey mathmari!

We can do it with trigonometry as well.
The width of the rectangle is $2\cos\phi$ and the height is $\sin\phi$, isn't it? (Thinking)
So the area is:
$$2\cos\phi \cdot \sin\phi = \sin(2\phi) =1\implies \phi=\frac\pi 4$$

Let $x$ be the height and $w$ the width. Since $M$ is the midpoint we get that $w=2y$.
At the right triangle we can Pythagoras' Theorem and we get that $y=\sqrt{1-x^2}$.
The area of the big rectangle is $1$ so we get that $x\cdot w=1 \Rightarrow x\cdot 2\sqrt{1-x^2}=1$ and from that eauation we can calculate $x$.

Btw we would get the same result if we would consider the acute angles to be $45^{\circ}$, so in this case they are indeed like that.

Yep. That works as well. (Nod)

 

[mathmari]

Thanks a lot!