Recall that the cosine of an angle is the ratio of the adjacent side to the hypotenuse of that triangle. Thus, for this triangle, we can say:

$\displaystyle \small cos(9.75)=\frac{x}{375}$

Solving for $\displaystyle \small x$, we get:

$\displaystyle \small 375*cos(9.75)=x$

$\displaystyle \small x=369.58352214677914$ or $\displaystyle \small 369.58$

Begin by drawing out this scenario using a little right triangle:

We know that the cosine of an angle is equal to the ratio of the side adjacent to that angle to the hypotenuse of the triangle. Thus, for our triangle, we know:

$\displaystyle cos(14.5) =\frac{x}{40}$

Using your calculator, solve for $\displaystyle x$:

$\displaystyle x=40cos(14.5)$

This is $\displaystyle 38.72590561512431$. Now, take the decimal portion in order to find the number of inches involved.

$\displaystyle 0.72590561512431 * 12 = 8.71086738149172$

Thus, rounded, your answer is $\displaystyle 38$ feet and $\displaystyle 9$ inches.

Use SOH-CAH-TOA to solve for the sine of a given angle. This stands for:

$\displaystyle \sin = \frac{\textup{opposite}}{\textup{hypotenuse}}$

$\displaystyle \cos = \frac{\textup{adjacent}}{\textup{hypotenuse}}$

$\displaystyle \tan = \frac{\textup{opposite}}{\textup{adjacent}}$.

From our triangle we see that at point $\displaystyle A$, the adjacent side is side $\displaystyle b$ and the hypotenuse doesn't depend upon position, it's always $\displaystyle c$. Thus we get that $\displaystyle cos(A)=\frac{b}{c}$

In right triangles, SOHCAHTOA tells us that $\displaystyle \cos A =\frac{\textup{adjacent}}{\textup{hypotenuse}}$, and we know that $\displaystyle \angle A = 42^{\circ}$ and hypotenuse $\displaystyle AC = 25$. Therefore, a simple substitution and some algebra gives us our answer.

$\displaystyle \cos 42^{\circ} = \frac{AB}{25}$

$\displaystyle .7 = \frac{AB}{25}$ Use a calculator or reference to approximate cosine.

$\displaystyle 17.5= AB$ Isolate the variable term.

Thus, $\displaystyle 17.5= AB$.

In right triangles, SOHCAHTOA tells us that $\displaystyle \cos C = \frac{\textup{adjacent}}{\textup{hypotenuse}}$, and we know that $\displaystyle A = 75^{\circ}$ and hypotenuse $\displaystyle AC = 42$. Therefore, a simple substitution and some algebra gives us our answer.

$\displaystyle \cos 75^{\circ} = \frac{CB}{42}$

$\displaystyle .3 = \frac{CB}{42}$ Use a calculator or reference to approximate cosine.

$\displaystyle 12.6= CB$ Isolate the variable term.

Thus, $\displaystyle 12.6= CB$.

The plane itself is effectively at the top of a right triangle, with topmost angle $\displaystyle 75^{\circ}$ and hypotenuse $\displaystyle 43$ feet. If this is the case, then SOHCAHTOA tells us that $\displaystyle \textup{cos} (75)^{\circ} = \frac{\textup{adjacent}}{\textup{hypotenuse}} = \frac{x}{43}$.

Now, solve for the adjacent:

$\displaystyle x = \textup{cos}(75^{\circ}) \cdot 43 \approx 11.13$

Thus, our plane's nose is approximately $\displaystyle 11.13$ feet from the runway.

Edgar is standing on top of a right triangle because the angle from the vertical ladder to the ground is 90 degrees. To solve this question, you must know SOHCAHTOA. This acronym can be broken into three parts to solve for the sine, cosine, and tangent.

$\displaystyle \textup{SOH: }\textup{Sine}=\frac{\textup{opposite}}{\textup{hypotenuse}}$

$\displaystyle \textup{CAH: }\textup{Cosine}=\frac{\textup{adjacent}}{\textup{hypotenuse}}$

$\displaystyle \textup{TOA: }\textup{Tangent}=\frac{\textup{opposite}}{\textup{adjacent}}$

In order to solve for the missing side, you need to choose the trigonometric function that includes the side you need to find and the side that you know, relative to the angle that you know. In this case, you know the hypotenuse, so you would not use the tangent function; furthermore, you are looking for the side that is adjacent to the 68-degree angle. Thus, you need the function that incorporates adjacent and hypotenuse—the cosine function.

$\displaystyle \cos 68^{\circ}=\frac{x}{35}$

$\displaystyle 35\cos 68^{\circ}=\frac{x}{35}\cdot 35$

$\displaystyle 35\cos 68^{\circ}=x$

Typically, you would use a calculator at this point to calculate the cosine function; however, based on the answer choices provided, you can stop at this point.

In right triangles, SOHCAHTOA tells us that $\displaystyle \cos A = \frac{\textup{adjacent}}{\textup{hypotenuse}}$, and we know that $\displaystyle A = 66^{\circ}$ and hypotenuse $\displaystyle AC = 8$. Therefore, a simple substitution and some algebra gives us our answer.

$\displaystyle \cos 66^{\circ} = \frac{AB}{8}$

$\displaystyle .41 = \frac{AB}{8}$ Use a calculator or reference to approximate cosine.

$\displaystyle 3.28= AB$ Isolate the variable term.

Thus, $\displaystyle 3.28= AB$.

The domain of a function is referring to the x values that can be plugged into the function and produce a value.

The domain of the parent function $\displaystyle cos(\theta)$ has a domain from negative infinity to positive infinity.  

The $\displaystyle -3$ term only shifts the function down three units, which will not affect the domain of the cosine graph.

Therefore, the answer is $\displaystyle {}(-\infty,\infty)$.

The function $\displaystyle y=2cos(\theta)+6$ is related to the parent function $\displaystyle y=cos(\theta)$.

The domain of the parent function is $\displaystyle (-\infty,\infty)$.  The values $\displaystyle 2$ and $\displaystyle 6$ will not affect the domain of the curve. 

The answer is $\displaystyle (-\infty,\infty)$.