First, solve for side MN. Tan(30^°) = MN/16√3, so MN = tan(30^°)(16√3) = 16. Triangle LMN and MNO are similar as they're both 30-60-90 triangles, so we can set up the proportion LM/MN = MN/NO or 16√3/16 = 16/x. Solving for x, we get 9.24, so the closest whole number is 9.

The angle of depression is the angle formed by a horizontal line and the line of sight looking down from the horizontal.

This is a right triangle trig problem. The vertical distance is 250 ft and the horizontal distance is unknown. The angle of depression is 35°. We have an angle and two legs, so we use tan Θ = opposite ÷ adjacent. This gives an equation of tan 35° = 250/d where d is the unknown distance to be directly over the house.

To solve this equation, it is best to remember the mnemonic SOHCAHTOA which translates to Sin = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent. Looking at the problem statement, we are given an angle and the side opposite of the angle, and we are looking for the side adjacent to the angle. Therefore, we will be using the TOA part of the mnemonic. Inserting the values given in the problem statement, we can write $\displaystyle TAN\left ( 55\right ) = \frac{9}{x}$. Rearranging, we get $\displaystyle x = \frac{9}{TAN\left ( 55\right )}$. Therefore $\displaystyle x = 6.3$

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

Note importantly: We are looking for $\displaystyle \small x$ as the the distance to the bottom of the building. Now, this is not very hard at all! We know that the tangent of an angle is equal to the ratio of the side adjacent to that angle to the opposite side of the triangle. Thus, for our triangle, we know:

$\displaystyle \small tan(37) =\frac{30}{x}$

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

$\displaystyle \small x=\frac{30}{tan(37)}$

This is $\displaystyle \small 39.8113446486123$. Now, take the decimal portion in order to find the number of inches involved.

$\displaystyle \small 0.8113446486123 * 12 = 9.7361357833476$

Thus, rounded, your answer is $\displaystyle \small 39$ feet and $\displaystyle \small 10$ inches.

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

$\displaystyle \small tan(27.45)=\frac{x}{50.45}$

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

$\displaystyle \small 50.45*tan(27.45)=x$

$\displaystyle \small x=26.20667657491611$ or $\displaystyle \small 26.21$

First we need to find the value of $\displaystyle \tan(B)$. Use the mnemonic SOH-CAH-TOA which stands for:
$\displaystyle \sin = \frac{opposite}{hypotenuse}$

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

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

Now we see at point $\displaystyle B$ we are looking for the opposite and adjacent sides, which are $\displaystyle b$ and $\displaystyle a$ respectively.
Thus we get that 

$\displaystyle \tan(B) = \frac{b}{a}$ 

and plugging in our values and reducing yields:

$\displaystyle \frac{4}{3}$

In right triangles, SOHCAHTOA tells us that $\displaystyle \tan A = \frac{\textup{opposite}}{\textup{adjacent}}$, and we know that $\displaystyle \angle A = 61^{\circ}$ and hypotenuse $\displaystyle AC = 3.87$. Therefore, a simple substitution and some algebra gives us our answer.

$\displaystyle \tan 61^{\circ} = \frac{CB}{3.87}$

$\displaystyle 1.80 = \frac{CB}{3.87}$ Use a calculator or reference to approximate cosine.

$\displaystyle 6.97= CB$ Isolate the variable term.

Thus, $\displaystyle 6.97= CB$.

In right triangles, SOHCAHTOA tells us that $\displaystyle \tan A = \frac{\textup{opposite}}{\textup{adjacent}}$, and we know that $\displaystyle \angle A = 50.5^{\circ}$ and hypotenuse $\displaystyle AC = 125$. Therefore, a simple substitution and some algebra gives us our answer.

$\displaystyle \tan 50.5^{\circ} = \frac{CB}{125}$

$\displaystyle 1.2 = \frac{CB}{125}$ Use a calculator or reference to approximate cosine.

$\displaystyle 150= CB$ Isolate the variable term.

Thus, $\displaystyle 150= CB$.

In right triangles, SOHCAHTOA tells us that $\displaystyle \tan A = \frac{\textup{opposite}}{\textup{adjacent}}$, and we know that $\displaystyle \angle A = 27^{\circ}$ and leg$\displaystyle AB = 300$. Therefore, a simple substitution and some algebra gives us our answer.

$\displaystyle \tan 27^{\circ} = \frac{CB}{300}$

$\displaystyle .5 = \frac{CB}{300}$ Use a calculator or reference to approximate cosine.

$\displaystyle 150= CB$ Isolate the variable term. 

Thus, $\displaystyle 150= CB$.