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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

enter image description here

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  • $\begingroup$ If available, would you add a link to the source? $\endgroup$ Commented Dec 26, 2020 at 17:43
  • $\begingroup$ @CarlChristian I only have archive/library paper copies. Would a scanned image of the (Swedish) problem be OK? $\endgroup$
    – mf67
    Commented Dec 26, 2020 at 17:54
  • $\begingroup$ Absolutely. I would be happy to read it. $\endgroup$ Commented Dec 26, 2020 at 18:22
  • $\begingroup$ @CarlChristian My pleasure. I've added it to the original question. $\endgroup$
    – mf67
    Commented Dec 26, 2020 at 18:48
  • $\begingroup$ Thank you very much! $\endgroup$ Commented Dec 26, 2020 at 19:34

2 Answers 2

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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.

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  • $\begingroup$ Much better! Many thanks! $\endgroup$
    – mf67
    Commented Dec 26, 2020 at 17:11
  • $\begingroup$ 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. $\endgroup$ Commented Dec 27, 2020 at 2:51
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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.

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  • $\begingroup$ @Cleo beat me to it :( $\endgroup$
    – user700480
    Commented Dec 26, 2020 at 16:55
  • $\begingroup$ Both equally good and well explained methods! Thank you. $\endgroup$
    – mf67
    Commented Dec 26, 2020 at 17:12
  • $\begingroup$ @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!) $\endgroup$
    – user700480
    Commented Dec 26, 2020 at 17:13
  • $\begingroup$ OK! I'm not very familiar with the ‘rules’ on S.E. Does one accept the first of several equal answers? $\endgroup$
    – mf67
    Commented Dec 26, 2020 at 17:18
  • $\begingroup$ Thank you @StinkingBishop, mf67 You are nice people :) But your answer is more complete. $\endgroup$
    – user851031
    Commented Dec 26, 2020 at 17:19

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