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 \(\displaystyle 0.95\).