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

Therefore:

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

$\displaystyle \small \tan{x}=\frac{\sin{x}}{\cos{x}}$

$\displaystyle \small \small \tan{x}=\frac{0.4}{0.2}$

$\displaystyle \small \tan{x}=0.035$

$\displaystyle \small x=\tan^{-1}0.035$

$\displaystyle \small x=2$

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.

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

We need the Tan B. Which side lengths correspond to this ratio?  

$\displaystyle \tan=\frac{opp}{adj}$  

The tangent function has a period of $\displaystyle \pi$ units. That is,

$\displaystyle tan(x+n\pi)=tanx$

for all $\displaystyle n\in\mathbb{Z}$.

Since $\displaystyle \frac{4\pi}{3}=\frac{\pi}{3}+\pi$, we can rewrite the original expression $\displaystyle tan(\frac{4\pi}{3})$ as follows:

$\displaystyle tan(\frac{4\pi}{3})=tan(\frac{\pi}{3}+\pi)$

                 $\displaystyle =tan(\frac{\pi}{3})$

                 $\displaystyle =\frac{sin(\frac{\pi}{3})}{cos(\frac{\pi}{3})}$

                 $\displaystyle =\frac{(\frac{\sqrt{3}}{2})}{(\frac{1}{2})}$

                 $\displaystyle =\sqrt{3}$

Hence, 

$\displaystyle tan(\frac{4\pi}{3})=\sqrt{3}$

First, convert the given angle measure from radians to degrees:

$\displaystyle cos(\frac{5\pi}{3})=cos(300^{\circ})$

Next, recall that $\displaystyle 300^{\circ}$ lies in the fourth quadrant of the unit circle, wherein the cosine is positive. Furthermore, the reference angle of $\displaystyle 300^{\circ}$ is 

$\displaystyle 360^{\circ}-300^{\circ}=60^{\circ}$

Hence, all that is required is to recognize from these observations that 

$\displaystyle cos(\frac{5\pi}{3})=cos(300^{\circ}) =cos(60^{\circ})$,

which is $\displaystyle \frac{1}{2}$.

Therefore,

$\displaystyle cos(\frac{5\pi}{3})=\frac{1}{2}$

Because $\displaystyle \sin x$ is an odd function, we can rewrite the second term in the expression.

$\displaystyle 2\sin(-h)=-2\sin h$.

We now use a double-angle formula to expand the first term.

$\displaystyle \sin(2h)\sec h=2\sin h\cos h\sec h$.

Because they are reciprocals, $\displaystyle \cos h\sec h=1$.

$\displaystyle 2\sin h\cos h\sec h-2\sin h=2\sin h-2\sin h=0.$

Before plugging the function into the calculator make sure the mode of the calculator is set to degrees,

Plug in $\displaystyle \sin72^{\circ}$ which equals to $\displaystyle 0.95$.