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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?

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

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  • $\begingroup$ Label the sides of orange a,b and h, lab the green h, d,and e. Then a+b+h=h+d+e. Note h=root (a^2+b^2) and e = root (h^2+e^2)=root (a^2+b^2+e^2). Note a=cos 20*h, b=sin 20*h. So you have a lot to play with. $\endgroup$ – fleablood Feb 26 '17 at 3:29
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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 ...

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... 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.

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  • $\begingroup$ @ Blue, why does squaring $sin\alpha + cos\alpha - tan\beta$ only gives $$(sin\alpha+cos\alpha)^2-2tan\beta+tan^2\beta$$" $\endgroup$ – pi-π Feb 26 '17 at 3:38
  • $\begingroup$ @ Blue, how is $$\begin{align} 2\tan\beta (\sin\alpha + \cos\alpha) &= ( \sin\alpha + \cos\alpha )^2 - 1 \tag{4}\\$$? $\endgroup$ – pi-π Feb 26 '17 at 3:42
  • $\begingroup$ Whoops ... typo. It's correct in (4). ... and now I've made the correction. Sorry for the confusion. $\endgroup$ – Blue Feb 26 '17 at 3:48
  • $\begingroup$ @ Blue, how is $$2tan\beta (sin\alpha+ cos\alpha)=(sin\alpha+cos\alpha)^2-1$$. Please elaborate $\endgroup$ – pi-π Feb 26 '17 at 4:01
  • $\begingroup$ In equation $(3)$, the $\tan^2\beta$ terms cancel. Then, if we subtract $1$ from both sides, and also add $2\tan\beta(\sin\alpha+\cos\beta)$ to both sides, we get equation $(4)$ (although the sides are reversed). $\endgroup$ – Blue Feb 26 '17 at 4:05
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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, $$x=\frac{Z^2+1}{2Z}$$

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  • $\begingroup$ @ ajotatxe, what is $x$? $\endgroup$ – pi-π Feb 26 '17 at 2:56
  • $\begingroup$ The first line of the answer says what is $x$. $\endgroup$ – ajotatxe Feb 26 '17 at 3:00
  • $\begingroup$ @ ajotatxe, where does $1$ come from? $\endgroup$ – pi-π Feb 26 '17 at 3:01
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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.

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  • $\begingroup$ My principal motivation in adding this solution is to show why it's often easier to let an arbitrary length be equal to $1$ (instead of a variable) in problems where scale doesn't matter. $\endgroup$ – Deepak Feb 26 '17 at 3:21
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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 $$tan \theta = \frac{\sqrt{1 - cos^2 \theta}}{cos \theta}$$. $$\frac{x_o}{h_o} = sin 20$$. $$\frac{y_o}{h_o} = cos 20$$.

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  • $\begingroup$ @ Srini., how do I get the exact angles? $\endgroup$ – pi-π Feb 26 '17 at 3:06
  • $\begingroup$ if $cos \theta = x$, by solving the equation $\frac{\sqrt{1 - x^2}}{x} = sin 20 + cos 20$ $\endgroup$ – Srini Feb 26 '17 at 3:14
  • $\begingroup$ solve for $x$ and then $\theta = cos^{-1} x$ $\endgroup$ – Srini Feb 26 '17 at 3:15

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