Two right triangles of equal perimeter share a side. Given an acute angle of one triangle, find the acute angles of the other.

Question

Two right triangles shown below have equal perimeters. The hypotenuse of the orange triangle is one leg of the green triangle stacked on top of it. If the smallest angle of the orange triangle is $20^\circ $, what are the angles of the green triangle?

Please help me. I didn't get any idea.

Answer 1

I'll work through a slightly-more-general problem, replacing $20^\circ$ with $\alpha$. Writing $d$ for the hypotenuse of the lower triangle, and $\beta$ for the angle shown in the upper triangle ...

... the equal perimeters condition tells us:

$$d + d \sin\alpha + d \cos\alpha = d + d\tan\beta + d\sec\beta \tag{1}$$

That is,

$$\sin\alpha + \cos\alpha - \tan\beta = \sec\beta \tag{2}$$

Squaring both sides of (2), and noting the $\sec^2\theta = \tan^2\theta + 1$, we have $$( \sin\alpha + \cos\alpha )^2 - 2 \tan\beta ( \sin\alpha + \cos\alpha ) + \tan^2\beta = \tan^2\beta + 1 \tag{3}$$ so that $$\begin{align} 2\tan\beta (\sin\alpha + \cos\alpha) &= ( \sin\alpha + \cos\alpha )^2 - 1 \tag{4}\\ &= \sin^2\alpha + \cos^2\alpha + 2 \sin\alpha \cos\alpha - 1 \tag{5}\\ &= 1 + 2\sin\alpha\cos\alpha - 1 \tag{6}\\ &= 2\sin\alpha\cos\alpha \tag{7} \end{align}$$

Therefore,

$$\tan\beta = \frac{\sin\alpha\cos\alpha}{\sin\alpha + \cos\alpha} \tag{$\star$}$$

In the particular case that $\alpha = 20^\circ$, we have $$\tan\beta = 0.250753\ldots \quad\to\quad \beta = \operatorname{atan} 0.25073\ldots = 14.0769\ldots^\circ$$

I don't believe there's a "nice" form for this angle.

Answer 2

If $a$ is the hypotenuse of the orange triangle, $\alpha$ is the least angle of the green triangle and $x=\cos\alpha$, $$a(1+\sin 20^\circ+\cos 20^\circ)=a(1+\tan\alpha+\sec\alpha)$$ $$\frac{1+\sqrt{1-x^2}}x=\sin20^\circ+\cos20^\circ$$

Now, if $Z=\sin 20^\circ+\cos 20^\circ$, $$1-x^2=Z^2x^2-2Zx+1$$ Then, since $x$ is not zero, $$x=\frac{Z^2+1}{2Z}$$

Answer 3

Let $x$ be the vertical sides, $y$ be the bases and $h$ be the hypotenuses. The subscripts $o$ for orange and $g$ for green. Let $\theta$ be the angle adjoining $20^o$ in the green triangle. $$tan \theta = \frac{x_g}{h_o}$$. $$cos \theta = \frac{h_o}{h_g}$$. $$x_g + h_g = x_o + y_o \ \text{(due to same perimeter)}$$. $$h_o tan \theta + \frac {h_o}{cos \theta} = x_o + y_o$$ and $$tan \theta = \frac{\sqrt{1 - cos^2 \theta}}{cos \theta}$$. $$\frac{x_o}{h_o} = sin 20$$. $$\frac{y_o}{h_o} = cos 20$$.

Answer 4

For simplicity, you can let the hypotenuse of the orange triangle be $1$ (it's just a matter of scale).

The perimeter of the orange triangle is $\sin 20^\circ + \cos 20^\circ + 1$

Letting the small angle of the green triangle be $\theta$, its perimeter is $\tan \theta + \sec \theta + 1$

Equating the two, you need to solve $\tan \theta + \sec \theta = \sin 20^\circ + \cos 20^\circ$

You can simplify the LHS by expressing it as a single trig ratio. $\sin 20^\circ + \cos 20^\circ = \sqrt 2\sin(45^\circ + 20^\circ) = \sqrt 2 \sin 65^\circ$

You can proceed as follows. Let $c = \cos\theta$

The equation can be rewritten:

$\frac{\sqrt{1-c^2} + 1}{c} = \sqrt 2\sin 65^\circ$

All that's left is to solve the quadratic in $c$ (easy as the constant term vanishes) then take the arccosine. The other angles are the ones complementary to this and the right angle. You should get (approx) $14.08^\circ$ for the small angle and $75.92^\circ$ for the complementary angle.