Elementary trigonometric proof problem using multiple angles in the sine rule

Question

In this diagram $AB=BC=CD=DE=1$meter,

Prove that $\angle AEB+\angle ADB = \angle ACB$

The method I used involved simple calculations using the sine rule and the pythagorean theorem, but I used a calculator, which isn't the most intuitive means of proving the statement, especially since the answer at the back of the book gave the following as a proof:

By the sine rule: $$\sin(180-\angle ADB - \angle AEB)=\frac{5\left(\frac{1}{\sqrt5}\right)}{\sqrt{10}}=\frac{1}{\sqrt2}$$ $$180-\angle ADB - \angle AEB = 135^{\circ}$$ $$\implies\angle ADB + \angle AEB = 45^{\circ}$$ and $$\sin(\angle ACB)=\frac{1}{\sqrt2}$$ $$\angle ACB = 45^{\circ}$$

I feel like there is something very obvious that I'm overlooking, but I have yet to find it. My main area of confusion is how does $\displaystyle \sin(180-\angle ADB - \angle AEB)=\frac{5\left(\frac{1}{\sqrt5}\right)}{\sqrt{10}}$.

Any help is appreciated, thank you.

FYI: The puzzle itself appears in the question "Show that the angles satisfy $x+y=z$", and my answer provides a trig-function-free solution. — Blue, Nov 04 '17 at 18:19

score 0 · Answer 1 · answered Nov 04 '17 at 18:25

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we have $$\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}=\frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\frac{1}{2}}=1=\tan(45^{\circ})$$

answered Nov 04 '17 at 18:25

Dr. Sonnhard Graubner

95,283

1

How does this clear up the area where my confusion lies? – joshuaheckroodt Nov 04 '17 at 18:26
you must use the tan-function, not sin – Dr. Sonnhard Graubner Nov 04 '17 at 18:27
I believe that, if OP were asking for a solution to the puzzle, then the question should be considered a duplicate of this one (and your solution effectively duplicates this answer). However, OP's specific concern is how a particular sine relation arises from the diagram. (Note: Your comment "you must use the tan-function" is an opinion apparently not shared by the author of OP's book.) – Blue Nov 04 '17 at 18:30
i don't know the author of OP's book – Dr. Sonnhard Graubner Nov 04 '17 at 18:31
i think this is the easiest way to solve this problem – Dr. Sonnhard Graubner Nov 04 '17 at 18:32
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@Dr.SonnhardGraubner: Of course you don't know the author. However, the solution in the book uses sine, and this is the source of the confusion for OP. Answers to the question should address that confusion, not offer alternative approaches to the puzzle (since these, as I mention, appear elsewhere on MSE). – Blue Nov 04 '17 at 18:33
this is your opinion, i think it is better to take an easy way – Dr. Sonnhard Graubner Nov 04 '17 at 18:34
i can add a proof by using the sin-function if you want – Dr. Sonnhard Graubner Nov 04 '17 at 18:38

Blue · Answer 2 · 2017-11-05T02:40:43.517

I had thought that "sine rule" meant "Law of Sines", but perhaps not. Consider this derivation of the relation at issue:

$$\begin{align} \sin(180^\circ-\angle ADB-\angle AEB) &= \sin(\angle ADB+\angle AEB) \tag{1}\\[4pt] &= \sin \angle ADB \cos \angle AEB + \cos \angle ADB \sin \angle AEB \tag{2}\\[4pt] &= \frac{|\overline{AB}|}{|\overline{AD}|}\frac{|\overline{AB}|}{|\overline{AE}|}+\frac{|\overline{BD}|}{|\overline{AD}|}\frac{|\overline{BE}|}{|\overline{AE}|} \tag{3}\\[4pt] &= \frac{1}{\sqrt{5}}\frac{3}{\sqrt{10}} + \frac{2}{\sqrt{5}} \frac{1}{\sqrt{10}} \tag{4}\\[4pt] &= \frac{5}{\sqrt{5}}\frac{1}{\sqrt{10}} \tag{5}\\[4pt] &= \frac{1}{\sqrt{2}} \tag{6} \end{align}$$ These steps are consistent with the appearance of "$5/\sqrt{5}$" in the book's proof, so it may be what the author intended. Even so, going from the sine expression to Step $(5)$ is quite a conceptual leap. Moreover, as shown in some answers to the puzzle as asked on MSE in 2016 (and in Dr. Graubner's solution here), a tangent-based derivation seems cleaner. (Of course, I believe a purely geometric approach such as the one I provided is cleaner still, but I'm biased.) All in all, it appears that the author could've done a much better job with the explanation.

I'll note that this puzzle appears on my Trigonography site, which includes a reference to both Numberphile's 2014 treatment, and Martin Gardner's 1996 Math Horizons article (which itself references a 1971 article by Charles Trigg(!), who found 54(!!) proofs of the relation).

Steven Alexis Gregory · Answer 3 · 2021-09-10T14:03:39.047

I want to give a trigonometric and a geometric solution.

Here is the trigonometric solution.

Clearly $\alpha = m\angle BCA = 45^\circ$

So we need to show that $\beta + \gamma = m\angle BDA + m\angle BEA = 45^\circ$

Clearly $0 < \beta, \gamma < 45^\circ$. So $0 < \beta + \gamma < 90^\circ$.

So, if we can show that $\tan(\beta+\gamma) = \tan 45^\circ = 1$, then it will follow that $\beta + \gamma = 45^\circ$

\begin{align} \tan(\beta+\gamma) &= \dfrac{\tan \beta + \tan \gamma}{1 + \tan \beta \cdot \tan \gamma} \\ &= \dfrac{\frac 12 + \frac 13}{1 - \frac 12 \cdot \frac 13} \\ &= \dfrac{\frac 56}{1 - \frac 16} \\ &= 1 \end{align}

Here is the geometric solution.

$\triangle ABE \sim \triangle AHG$

$\triangle ABD \sim \triangle GFA$

$m\angle HAF = 45^\circ = \alpha$

$\therefore \alpha + \beta = \gamma$.

Elementary trigonometric proof problem using multiple angles in the sine rule

3 Answers3

Here is the trigonometric solution.

Here is the geometric solution.