5

The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.

The sum of two angles are $135^\circ$ and the sum of their tangents are $5$. Calculate the angles.

Is there a shorter/simpler solution than the one presented below that I made some months ago? It seems rather ‘lengthy’.

Solution

Let the angles be $\alpha$ and $\beta$.

We have $$ \left\{ \begin{aligned} \alpha+\beta&=135°,\\ \tan(\alpha)+\tan(\beta)&=5. \end{aligned} \right. $$

Since $$ \tan(x)+\tan(y) =\frac{\sin(x)}{\cos(x)}+\frac{\sin(y)}{\cos(y)} =\frac{\sin(x)\cos(y)+\cos(x)\sin(y)}{\cos(x)\cos(y)} =\frac{\sin(x+y)}{\cos(x)\cos(y)} $$ we have $$ 5 =\tan(\alpha)+\tan(\beta) =\frac{\sin(\alpha+\beta)}{\cos(\alpha)\cos(\beta)} =\frac{\sin(135°)}{\cos(\alpha)\cos(\beta)} =\frac{\frac{1}{\sqrt{2}}}{\cos(\alpha)\cos(\beta)} $$ which gives $$ \cos(\alpha)\cos(\beta)=\tfrac{1}{5\sqrt{2}}. $$

Further $$ -\tfrac{1}{\sqrt{2}} =\cos(135°) =\cos(\alpha+\beta) =\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta) =\tfrac{1}{5\sqrt{2}}-\sin(\alpha)\sin(\beta) $$ which gives $$ \sin(\alpha)\sin(\beta) =\tfrac{1}{5\sqrt{2}}+\tfrac{1}{\sqrt{2}} =\tfrac{6}{5\sqrt{2}}. $$

Since $\alpha+\beta=135°$ we have \begin{align*} \sin(\alpha)\sin(\beta) & =\sin(\alpha)\sin(135°-\alpha) \\&=\sin(\alpha)\bigl(\sin(135°)\cos(\alpha)-\cos(135°)\sin(\alpha)\bigr) \\&=\sin(\alpha)\bigl(\tfrac{1}{\sqrt{2}}\cos(\alpha)+\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigr) \\&=\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigl(\cos(\alpha)+\sin(\alpha)\bigr) \\&=\tfrac{1}{\sqrt{2}}\bigl(\sin(\alpha)\cos(\alpha)+\sin^2(\alpha)\bigr) \\&=\tfrac{1}{\sqrt{2}}\Bigl(\tfrac{1}{2}\sin(2\alpha)+\tfrac{1}{2}\bigl(1-\cos(2\alpha)\bigr)\Bigr) \\&=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr) \end{align*} why \begin{gather*} \tfrac{6}{5\sqrt{2}}=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr) \\\quad\Leftrightarrow\quad \tfrac{12}{5}=\sin(2\alpha)+1-\cos(2\alpha) \\\quad\Leftrightarrow\quad \tfrac{7}{5}=\sin(2\alpha)-\cos(2\alpha) \\\quad\Leftrightarrow\quad \tfrac{7}{5}=\sqrt{2}\sin(2\alpha-\tfrac{\pi}{4}) \\\quad\Leftrightarrow\quad \tfrac{7}{5\sqrt{2}}=\sin(2\alpha-\tfrac{\pi}{4}) \end{gather*} which gives \begin{gather*} 2\alpha-\tfrac{\pi}{4}= \begin{cases} \arcsin(\tfrac{7}{5\sqrt{2}})+n_12\pi\\ \pi-\arcsin(\tfrac{7}{5\sqrt{2}})+n_22\pi \end{cases} \\\quad\Leftrightarrow\quad \alpha= \begin{cases} \tfrac{1}{2}\bigl(\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_12\pi\bigr)\\ \tfrac{1}{2}\bigl(\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_22\pi\bigr) \end{cases} \\\quad\Leftrightarrow\quad \alpha= \begin{cases} \tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{8}+n_1\pi\\ -\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{5\pi}{8}+n_2\pi \end{cases} \end{gather*} where $\beta=135°-\alpha=\frac{3\pi}{4}-\alpha$ and $n_1,n_2\in\mathbb{Z}$, and vice versa since the problem is ‘symmetrical’.

The original exam

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N. F. Taussig
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mf67
  • 853

2 Answers2

13

Take $\tan$ of both sides of first equation, $$\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta}$$

$$\Rightarrow \tan 135 = -1 = \dfrac{5}{1-\tan \alpha \tan \beta} $$

$$\Rightarrow \tan \alpha \tan \beta=6$$

Setting $\tan \alpha = x$, $\tan \beta = y$ we get two equations : $$x+y=5 \quad xy=6$$

I believe you can easily solve from here.

  • Much better! Many thanks! – mf67 Dec 26 '20 at 17:11
  • One could also calculate $\tan\beta=\tan(135-\alpha)$ using the tangent-addition formula and reduce everything to an equation in $\tan\alpha$. I like the approach in the answer better, but this approach might possibly be easier to stumble upon. – Greg Martin Dec 27 '20 at 02:51
5

Recall that $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$. Now, given $\alpha+\beta=135^{\circ}$ and $\tan(\alpha+\beta)=\tan 135^\circ=-1$, you get $\frac{5}{1-\tan\alpha\tan\beta}=-1$, i.e. $\tan\alpha\tan\beta=6$. Now, knowing the sum ($5$) and the product ($6$) of $\tan\alpha$ and $\tan\beta$, using Vieta formulas we conclude that $\tan\alpha$ and $\tan\beta$ are the solutions of the quadratic equation $x^2-5x+6=0$, i.e. $\tan\alpha$ and $\tan\beta$ are $2$ and $3$ in some order. This is now very easy to finish off: $\alpha=\arctan 2+n\cdot 180^{\circ}$ and $\beta=135^{\circ}-\alpha=\arctan 3-n\cdot 180^{\circ}$, or vice versa.

  • @Cleo beat me to it :( –  Dec 26 '20 at 16:55
  • Both equally good and well explained methods! Thank you. – mf67 Dec 26 '20 at 17:12
  • @mf67 Sadly you can only accept one answer, so I suggest you accept the quicker one. (I don't mind, and the idea is exactly the same!) –  Dec 26 '20 at 17:13
  • OK! I'm not very familiar with the ‘rules’ on S.E. Does one accept the first of several equal answers? – mf67 Dec 26 '20 at 17:18
  • Thank you @StinkingBishop, mf67 You are nice people :) But your answer is more complete. –  Dec 26 '20 at 17:19
  • You can accept any answer you like. The author of that answer gets additional reputation. As this is a reference site, people will look up to accepted answers in the future, so it is a good idea to accept a correct answer that actually answers the question. Apart from that, it is up to you. You can also see the number of upvotes, but this only tells you whether other visitors thought the answer was good - it is not telling you what you should think about answers. As it happens, popular opinion is also not always right... –  Dec 26 '20 at 17:23