What is $\tan50^\circ$? (without using a calculator)
I think the answer is 3, but I can not explain this mathematically. The only logic I came up with is that $45^\circ<50^\circ<60^\circ$ and therefore $\tan45^\circ<\tan50^\circ<\tan60^\circ$; that is $1<\tan50^\circ<\sqrt{3}$.
Is there a better approach to this problem?
Your intuition is correct. If you want to be more precise, observe that $$\tan(50) = \tan(45 + 5) = {\tan(45) + \tan(5) \over 1 - \tan(45)\tan(5)}$$ $$= {1 + \tan(5) \over 1 - \tan(5)}$$ Since $\tan(x)$ is very close to ${\pi \over 180} x$ for small $x$, this should be very close to $${1 + {\pi \over 36} \over 1 - {\pi \over 36}}$$ This is equal to $1.1912..$, while the true value is $1.1917..$.
The $\tan$ function is growing on $\left[0,\frac{\pi}2\right[$
We also know that $\tan(\frac{\pi}4)=1$ and $50°>\frac{\pi}4=45°$
What is more around $\frac{\pi}4$ the function has a slope of about 1, therefore a $5°$ difference won't induce a big difference between $\tan(45°)$ and $\tan(50°)$.
Therefore $\tan(50°)$ is a little bigger than $1$ which is answer $3$.
Using the series approximation of cotangent, $\cot{x}\approx\frac{1}{x}-\frac{x}{3}$,
$$\begin{align} \tan{\frac{5\pi}{18}}&=\cot{\frac{2\pi}{9}}\\ &\approx \left(\frac{2\pi}{9}\right)^{-1}-\frac13\left(\frac{2\pi}{9}\right)\\ &\approx 1.19968392... \end{align}$$
Note that this simple two term approximation differs from the true value by less than $0.7\%$.
Let me work with radians and let us have a look to $\tan(\frac{5\pi}{18})$. We can write $$\tan(\frac{5\pi}{18})=\tan(\frac{\pi}{4}+\frac{\pi}{36})$$ At this point, we could apply the formula for $\tan(a+b)$ and conclude. Another way is to develop $\tan(x)$ as a Taylor series at $x=\frac{\pi}{4}$ and get $$\tan(x)=1+2 \left(x-\frac{\pi }{4}\right)+O\left(\left(x-\frac{\pi }{4}\right)^2\right)$$ If we use $x=\frac{\pi}{4}+\frac{\pi}{36}$, we then have $$\tan(\frac{5\pi}{18}) \simeq 1+\frac{\pi}{18} \simeq 1.17453$$ while the exact value is $1.19175$.
Another solution is to draw the circle, the line corresponding to the two angles and observe where are located the values of the respective tangents.
I would draw a $45^\circ-45^\circ-90^\circ$ triangle (tangent $45^\circ$ IS actually 1, of course) and observe the effect on the tangent ratio of increasing one of the $45^\circ$ angles while preserving the $90^\circ$ angle.
Note that
$$\tan(45^o) < \tan(50^o) < \tan(\frac{45^o+60^o}{2})$$
Now, using the formula (which can be deduced easily by writing $\tan=\frac{\sin}{\cos}$) $$\tan(\frac{A+B}{2})=\frac{\sin(A)+\sin(B)}{\cos(A)+\cos(B)}$$ we get $$\tan(\frac{45^o+60^o}{2})=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+1}=(\sqrt{2}+\sqrt{3})(\sqrt{2}-1) \sim 1.3$$
Is there a better approach to this problem?
No. The only thing needed is mentioning that $\tan x$ is strictly increasing on $\bigg[- \dfrac\pi2,\dfrac\pi2\bigg]$, which can easily be proven either by drawing the trigonometric circle, or by using calculus, since $\tan'x$ $=1+\tan^2x>0$.
The addition formula for $\tan$ is
$$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$$
Let $A=45^{\circ}$ and $B = 5^{\circ}$. This gives us $$\tan 50^{\circ} = \frac{\tan 45 + \tan 5}{1-\tan 45 \tan 5}$$ We know that $\tan 45 = 1$ and so
$$\tan 50 = \frac{1+\tan 5}{1-\tan 5}$$
Imagine a right-angled triangle with a $5^{\circ}$ angle. The adjacent side would be much, much longer than the opposite side: $0 < \tan 5 < 1$. It is clear that $\tan 50$ is very close to $1$. But is it bigger than or smaller than $1$? Well, since $0 < \tan 5 < 1$ we have $1+\tan > 1-\tan 5$ and so $$\frac{1+\tan 5}{1-\tan 5} > 1$$
The answer is (c).
