Can I find the length of the side of a triangle WITHOUT a calculator? 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?
 A: When you calculated it with your scientific calculator, he interpreted the angle (65) as radians.


*

*$5 \cdot \tan(65) = -7.35$ (when using rad) (Wolfram Alpha)

*$5 \cdot \tan(65) = 10.72$ (when using deg) (Wolfram Alpha) 


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)
A: 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 $\pi/36\approx 0.09$.
A: 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.
A: 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
A: 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.
A: $$5 \cdot \tan(65°) \simeq 10.7225 $$
$$5 \cdot \tan(65) \simeq    -7.3502 $$
A: 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). 

A: 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:
$$
\begin{cases}
\sin\left(65^{\circ}\right)=\frac{a}{b}\\
\cos\left(65^{\circ}\right)=\frac{5}{b}\\
\tan\left(65^{\circ}\right)=\frac{a}{5}
\end{cases}\to a=5\tan\left(65^{\circ}\right)\space\space\text{and}\space\space b=\frac{5}{\cos\left(65^{\circ}\right)}
$$
A: Did you already consider a slide rule? (See Wikipedia: https://en.wikipedia.org/wiki/Slide_rule)
