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I was on Khan Academy watching a video where he was finding the length of a side of a right triangle where one angle measured 65° and the length of the side adjacent to the angle was 5. The side he was trying to find was opposite to the angle that measured 65°, so to find the length of the side, he used $tan(65°) = \frac{a}{b}$, and he got 10.7 when he multiplied 5 by $tan(65°)$.

(This is the link to the video: https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/trig-solve-for-a-side/v/example-trig-to-solve-the-sides-and-angles-of-a-right-triangle)

Since I don't have a graphing calculator, I used my scientific calculator to do the same thing, but I got 7.8, and I also used the calculator on my phone which gave me -7.4. Is there a way I can do the same thing WITHOUT a calculator?

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    $\begingroup$ are you sure that $65$ are degrees and not radians for your scientific calculator? $\endgroup$
    – the_candyman
    Aug 25, 2016 at 19:59
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    $\begingroup$ You certainly don't need a graphing calculator to get the tangent of an angle. Your calculator should do fine. I think @the_candyman is spot on with his assessment of the issue here. $\endgroup$
    – Klint Qinami
    Aug 25, 2016 at 20:13
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    $\begingroup$ Unfortunately for you, there is no nice way. For future related $tan$ problems, you could note that $tan(\theta) = sin(\theta) / cos(\theta)$ if you already know the sine and cosine of the angle. $\endgroup$
    – Tristan Batchler
    Aug 25, 2016 at 20:15
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    $\begingroup$ -7.4 is consistent with $5 \tan 65$ and the calculator setting in radians. As far as doing trig without a calculator... very tough to do this sort of question without a calculator. You can get a slide rule. Or a protractor and a ruler. $\endgroup$
    – Doug M
    Aug 25, 2016 at 20:20

9 Answers 9

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When you calculated it with your scientific calculator, he interpreted the angle (65) as radians.

You don't need a graphic calculator for this sort of problem, but without any calculator you will have answers like those given in the video:

$$a = 5 \cdot \tan(65°)$$ $$b = \frac{5}{\cos(65°)}$$ There is nothing wrong with those, they just aren't calculated to numbers (but they would get rounded anyway, so many might prefere this form)

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In this case, since $65^\circ \approx 60^\circ$, Use the fact that $\tan 60^\circ = \sqrt{3}$.

Now use the fact that $\tan x \approx x$ for small angles, and that: $$\tan(a+b)=\frac{\tan a + \tan b}{1-\tan a\tan b}$$ So we have that: $$\tan 65^\circ = \frac{\tan 60^\circ + \tan 5^\circ}{1-\tan 60^\circ\tan 5^\circ}$$ Now, $$\tan 5^\circ = \tan 5\pi/180=\tan \pi/36 \approx \tan 0.1 \approx 0.1$$ So that using the known value $\sqrt{3}\approx 1.73$, we have: $$\tan 65^\circ \approx \frac{1.73 + 0.1}{1-0.173}\approx2.21$$ Which is only $3\%$ off the true value. You could get even better by using $\pi/36\approx 0.09$.

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I don't know where the $7.8$ came from with your scientific calculator, but the calculator in your phone was likely in radians and not degrees.

$$ 5 \tan (65^\circ) \approx 10.72$$

and

$$ 5 \tan(65 \text{ rad}) \approx -7.35$$

Is there a way I can do the same thing WITHOUT a calculator?

No nice way that I can think of. $65 = 60 + 5$ and $\tan 60^\circ$ is trivial to work with, but $\tan 5^\circ$ is not, as far as I know. If there's a way to express $65^\circ$ as an expression involving angles (and possibly constant multiples) that play "nicely" with tangent, then yes. But I don't think such an expression exists.

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    $\begingroup$ Since we're talking about approximation, not an exact value, then for $\tan 5^{\circ}$, that angle is small enough we can convert to radians and use the small-angle approximation for tangent to estimate $\tan 5^{\circ}$. In this case that gives $\tan 5^{\circ} \approx 0.087266$, while the actual value is closer to 0.087489, thus good to in under 1%. $\endgroup$
    – The_Sympathizer
    Aug 25, 2016 at 21:58
  • $\begingroup$ Well you could do it with trig tables, but that's merely a literal answer to the question, since there's no reason why you would unless you didn't have a calculator, but did have a tan table, which seems ... unlikely these days $\endgroup$
    – Au101
    Aug 26, 2016 at 2:58
  • $\begingroup$ I prefer using a slide rule than a trig table - as I have two of the former (still beside my desk) and only one of the latter; buried in my basement bookshelf. That give me about 10.2 +- 0.1 $\endgroup$
    – Pieter Geerkens
    Aug 26, 2016 at 6:14
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The answer is obviously that you used radians instead of degrees.

If you really want to calculate arbitrary tangent without a calculator,one way to go is to first convert to radians by multiplying your angle by $\frac{\pi}{180}$. Then pretend you're a calculator by computing

$$\sin \theta = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$

and

$$\cos \theta = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$$

Obviously don't go to infinity or else you'll be calculating for a long time, but use however much precision you need. When you have the two values, your tangent will just be

$$\tan \theta = \frac{\sin \theta}{\cos \theta}$$

**Note that if you're far away from zero, these series will take longer to converge

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    $\begingroup$ "don't go to infinity or else you'll be calculating for a long time." - Understatement of the year. $\endgroup$
    – Brevan Ellefsen
    Aug 25, 2016 at 22:09
  • $\begingroup$ I suppose that already the approximation for $\pi$ will only have a limited precision. $\endgroup$
    – Hagen von Eitzen
    Aug 26, 2016 at 9:00
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I learned basic trig in the 1960's, before people carried around phones that can compute trig functions, so I naturally know how to do it without a calculator. The key tool is a book of tables of mathematical functions.

I looked up the tan(65o) in the "Natural Tangents" table in "Chambers Seven-Figure Mathematical Tables". It gave 2.1445069. Multiplying by 5, I get 10.7225345. If I only wanted one decimal place, I would report it as 10.7.

Most of the time to do the calculation was spent finding a book I have not used in several decades, but kept for sentimental reasons.

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$$5 \cdot \tan(65°) \simeq 10.7225 $$

$$5 \cdot \tan(65) \simeq -7.3502 $$

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The correct answer is: $$5\cdot \tan(65°) \approx 10.722 $$

You did one calculation as $$5\cdot \tan(65) \approx -7.35 \text{ (angle of 65 radians)}$$

The other you calculated as $$5\cdot \text{atan}(65)\approx 7.78 \text{ (arctan, in radians, so double wrong)}$$

Angles are commonly measured in radians where $$2\pi \text{ radians is 360°}$$ (and rarely in grads where 360° = 400 grads). Calculators often have modal settings for all three.

atan() is the inverse function to tan(), and will give you the angle from the sides. In the example, atan(10.722/5) ~= 65° (or 1.134 radians).

enter image description here

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What do we know:

  • Angle: $$\angle\text{W}+\angle\text{X}+\angle\text{Y}=180^{\circ}=\pi\space\text{radians}$$

We know that $\angle\text{X}=90^{\circ}=\frac{\pi}{2}\space\text{radians}$ and $\angle\text{Y}=65^{\circ}=\frac{13\pi}{36}\space\text{radians}$, so:

$$\angle\text{W}+90^{\circ}+65^{\circ}=180^{\circ}=\pi\space\text{radians}\Longleftrightarrow\angle\text{W}=25^{\circ}=\frac{5\pi}{36}\space\text{radians}$$

  • It is a right triangle so we can use the Pythagorean Theorem: $$\text{WY}^2=\text{XY}^2+\text{WX}^2$$

We know that sides $\text{XY}=5$, $a=\text{WX}$ and $b=\text{WY}$, so:

$$b^2=5^2+a^2\Longleftrightarrow b^2-a^2=25$$

  • Using sin/cos/tan in a right triangle, we can say:

$$ \begin{cases} \sin\left(\angle\text{Y}\right)=\frac{\text{WX}}{\text{WY}}\\ \cos\left(\angle\text{Y}\right)=\frac{\text{XY}}{\text{WY}}\\ \tan\left(\angle\text{Y}\right)=\frac{\text{WX}}{\text{XY}} \end{cases} $$

We know that sides $\text{XY}=5$, $a=\text{WX}$, $b=\text{WY}$ and $\angle\text{Y}=65^{\circ}=\frac{13\pi}{36}\space\text{radians}$, so:

$$ \begin{cases} \sin\left(65^{\circ}\right)=\frac{a}{b}\\ \cos\left(65^{\circ}\right)=\frac{5}{b}\\ \tan\left(65^{\circ}\right)=\frac{a}{5} \end{cases}\to a=5\tan\left(65^{\circ}\right)\space\space\text{and}\space\space b=\frac{5}{\cos\left(65^{\circ}\right)} $$

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Did you already consider a slide rule? (See Wikipedia: https://en.wikipedia.org/wiki/Slide_rule)

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