Two right angles with an acute angle of 17 degrees must be similar because triangles that are similar share the same angles.Is this proper?
[Math] Explain why two right triangles, each with an acute angle of 17 degrees, must be similar.
trianglestrigonometry
Related Solutions
Experimentally it appears that seems that something close to Hagen von Eitzen's comment of the construction using a regular pentagon
works for all obtuse triangles. Here is an example, with two isosceles triangles cutting off the sharp angles, and then choosing a suitable point inside the remaining pentagon to give $7$ acute angled triangles in all. Not all pairs of isosceles triangles allow there to be a suitable interior point, but I could not find an obtuse angled triangle without some solution.
I would be surprised if there was a solution with fewer acute angled triangles.
I think I have a solution but it will be difficult to explain. I will use the following variables:
$$ \begin{split} a&=\text{base of orange triangle} \\ b&=\text{height of orange triangle} \\ c&= \text{"left" side of green triangle} \end{split} $$
Then the hypotenuse of the orange triangle is $\sqrt{a^2+b^2}$ and the hypotenuse of the green triangle is $\sqrt{a^2+b^2+c^2}$. Since the perimeters of both triangles are equal we must have:
$$ \begin{split} &&a+b=c+\sqrt{a^2+b^2+c^2} \\ &\implies& a+b-c=\sqrt{a^2+b^2+c^2} \\ &\implies& (a+b-c)^2=a^2+b^2+c^2 \\ &\implies& a^2+b^2+c^2+2ab-2ac-2bc=a^2+b^2+c^2 \\ &\implies& ab=ac+bc \\ &\implies& c=\frac{ab}{a+b} \end{split} $$
So now we have all sides in terms of $a$ and $b$. But we can relate $a$ and $b$ using the angle we are given. So:
$$ \tan(20)=\frac{b}{a} \implies b=a\tan(20) $$
Using this we get:
$$ \begin{split} a^2+b^2=a^2+a^2\tan^2(20)=a^2(1+\tan^2(20))=\frac{a^2}{\sin^2(20)} \end{split} $$
Now we can get an expression for $a^2+b^2+c^2$ purely in terms of $a$:
$$ \begin{split} a^2+b^2+c^2&= \frac{a^2}{\sin^2(20)}+\frac{a^2b^2}{\frac{a^2}{\sin^2(20)}+2ab}\\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)}{\frac{a^2}{\sin^2(20)}+2a^2\tan(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^4\tan^2(20)\sin^2(20)}{a^2+2a^2\tan(20)\sin^2(20)} \\ &=\frac{a^2}{\sin^2(20)}+\frac{a^2\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \\ &=a^2\left[ \frac{1}{\sin^2(20)}+\frac{\tan^2(20)\sin^2(20)}{1+2\tan(20)\sin^2(20)} \right] \\ &= a^2K^2 \end{split} $$
where $K^2$ is just the number in the square brackets (the only reason I made it $K^2$ instead of $K$ is because we will need to take square roots). Finally, letting $\theta$ be the smallest angle of the green triangle, we get:
$$ \begin{split} \cos(\theta)&=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2+c^2}} \\ &=\frac{\frac{a}{\sin(20)}}{aK} \\ &=\frac{1}{K\sin(20)} \end{split} $$
You can work out that $K\approx 2.9262$ and then plugging this in gives us $\theta\approx 2.34^{\circ}$.
Best Answer
It's not that all pairs of similar triangles share the same angles; rather the essential fact is that all pairs of triangles that share the same angles are similar.
You say they are both right triangles. That means they both have a $90^\circ$ angle. They also both have a $17^\circ$ angle.
It follows that both triangles have a $73^\circ$ angle: $$ 180^\circ-90^\circ-17^\circ = 73^\circ. $$ So they have three angles in common.