First lets identify the different quadrants.

Quadrant I:$\displaystyle 0\rightarrow \frac{\pi}{2}$

Quadrant II: $\displaystyle \frac{\pi}{2}\rightarrow \pi$

Quadrant III: $\displaystyle \pi \rightarrow \frac{3\pi}{2}$

Quadrant IV: $\displaystyle \frac{3\pi}{2}\rightarrow 2\pi$

Now looking at our possible answer choices, we will add or subtract $\displaystyle 2\pi$ until we get the reduced fraction of the angle. This will tell us which quadrant the angle lies in.

$\displaystyle \small \frac{-2\pi}{3}+2\pi=\frac{-2\pi}{3}+\frac{6\pi}{3}=\frac{4\pi}{3}$ thus in quadrant III. 

$\displaystyle \small \frac{13\pi}{4}-2\pi=\frac{13\pi}{4}-\frac{8\pi}{4}=\frac{5\pi}{4}$ thus in quadrant III.

Therefore,

$\displaystyle \small \frac{-2\pi}{3}$ and $\displaystyle \small \frac{13\pi}{4}$ is the correct answer.

First lets identify the different quadrants.

Quadrant I:$\displaystyle 0\rightarrow \frac{\pi}{2}$

Quadrant II: $\displaystyle \frac{\pi}{2}\rightarrow \pi$

Quadrant III: $\displaystyle \pi \rightarrow \frac{3\pi}{2}$

Quadrant IV: $\displaystyle \frac{3\pi}{2}\rightarrow 2\pi$

The correct answer,$\displaystyle \small \frac{-4\pi}{3}$, is coterminal with $\displaystyle \small \frac{2\pi}{3}$.

We can figure this out by adding $\displaystyle \small 2\pi$, or equivalently $\displaystyle \small \frac{6\pi}{3}$ to get $\displaystyle \small \frac{2\pi}{3}$, or we can count thirds of pi around the unit circle clockwise. Either way, it is the only angle that ends in the second quadrant.

One way to uncover which quadrant this angle lies is to ask how many complete revolutions this angle makes by dividing it by 360 (and rounding down to the nearest whole number).

With a calculator we find that $\displaystyle 50000^o$ makes $\displaystyle 138$ full revolutions. Now the key lies in what the remainder the angle makes with $\displaystyle 138$ revolutions:

$\displaystyle 50000^o - 138 * 360^o = 320^o$

$\displaystyle 270^o < 320^o < 360^o$, therefore our angle lies in the fourth quadrant.

Alternatively, we could find evaluate $\displaystyle sin \ 50000^o$ and $\displaystyle cos \ 50000^o$.

The former (sine) gives us a negative number whereas the latter (cosine) gives a positive. The only quadrant in which sine is negative and cosine is positive is the fourth quadrant.

Step 1: Define the quadrants and the angles that go in:

QI:

$\displaystyle 0^\circ< x\leq 90^\circ$

QII:

$\displaystyle 90^\circ < x \leq 180^\circ$

QIII:

$\displaystyle 180^\circ < x \leq 270^\circ$

QIV:

$\displaystyle 270^\circ < x \leq 360$

Step 2: Find the quadrant where $\displaystyle 135^\circ$ is:

The angle is located in QII (Quadrant II)

First, using the unit circle, we can see that the denominator has a four in it, which means it is a multiple of $\displaystyle \pi/4$.

We want to reduce the angle down until we can visualize which quadrant it is in. You can subtract $\displaystyle 2\pi$ away from the angle each time (because that is just one revolution, and we end up at the same spot).

If you subtract away $\displaystyle 2\pi$ twice, you are left with $\displaystyle \pi/4$, which is in quadrant I. 

$\displaystyle 17\pi/4-8\pi/4-8\pi/4=\pi/4$.

The second quadrant contains angles between $\displaystyle \frac{\pi }{2}$ and $\displaystyle \pi$, plus those with additional multiples of $\displaystyle 2\pi$.  The angle $\displaystyle \frac{11\pi }{4}$ is, after subtracting $\displaystyle 2\pi$, is simply $\displaystyle \frac{3\pi }{4}$, which puts it in the second quadrant.

The cosine of an angle is the value of the adjacent side over the hypotenuse.

Therefore:

$\displaystyle cos \angle B = \frac{adjacent}{hypotenuse} = \frac{7}{25}$

Solve each term separately.

$\displaystyle cos(30)= \frac{\sqrt3}{2}$

$\displaystyle cos(60)=\frac{1}{2}$

Add both terms.

$\displaystyle \frac{\sqrt3}{2}+\frac{1}{2}= \frac{\sqrt3+1}{2}$

Rewrite $\displaystyle 2tan(120)$ in terms of sines and cosines.

$\displaystyle 2tan(120)=2(\frac{sin(120)}{cos(120)})=2(\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}})$

Simplify the complex fraction.

$\displaystyle 2(\frac{\frac{\sqrt3}{2}}{-\frac{1}{2}})= 2\times\frac{\sqrt3}{2}\times-2 = -2\sqrt3$

To find the value of $\displaystyle \frac{1}{2}sin(45)+ tan(60)$, solve each term separately.

$\displaystyle \frac{1}{2}sin(45)=\frac{1}{2} \cdot \frac{\sqrt2}{2} = \frac{\sqrt2}{4}$

$\displaystyle tan(60) = \sqrt3$

Sum the two terms.

$\displaystyle \frac{\sqrt2}{4}+\sqrt3 = \frac{\sqrt2}{4}+\frac{4\sqrt3}{4} = \frac{\sqrt2+4\sqrt3}{4}$