sin(30º) = ½

sine = opposite / hypotenuse

½ = opposite / 8

Opposite = 8 * ½ = 4

Use SOHCAHTOA. Sin(30) = 12/x, then 12/sin(30) = x = 24.

You can also determine the side with a measure of 12 is the smallest side in a 30:60:90 triangle. The hypotenuse would be twice the length of the smallest leg.

We can solve for the length of the chord by drawing a line the bisects the angle and the chord, shown below as $\displaystyle \overline{AD}$.

In this circle, we can see the triangle $\displaystyle \bigtriangleup ADC$ has a hypotenuse equal to the radius of the circle ($\displaystyle \overline{AC}$), an angle $\displaystyle \theta$ equal to half the angle made by the chord, and a side $\displaystyle \overline{CD}$ that is half the length of the chord.  By using the sine function, we can solve for $\displaystyle \overline{CD}$.

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

$\displaystyle \sin \left ( \theta\right ) = \frac{\overline{CD}}{\overline{AC}}$

$\displaystyle \sin \left ( 55^{\circ}\right ) = \frac{\overline{CD}}{5}$

$\displaystyle 5\sin \left ( 55^{\circ}\right ) = \overline{CD}$

$\displaystyle 4.1 = \overline{CD}$

The length of the entire chord is twice the length of $\displaystyle \overline{CD}$, so the entire chord length is $\displaystyle 8.2$.

We can solve for the length of the chord by drawing a line the bisects the angle and the chord, shown below as $\displaystyle \overline{AD}$.

In this circle, we can see the triangle $\displaystyle \bigtriangleup ADC$ has a hypotenuse equal to the radius of the circle ($\displaystyle \overline{AC}$), an angle $\displaystyle \theta$ equal to half the angle made by the chord, and a side $\displaystyle \overline{CD}$ that is half the length of the chord.  By using the sine function, we can solve for $\displaystyle \overline{CD}$.

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

$\displaystyle \sin \left (\theta\right ) = \frac{\overline{CD}}{\overline{AC}}$

$\displaystyle \sin \left ( 63.5^{\circ}\right ) = \frac{\overline{CD}}{8}$

$\displaystyle 8\sin \left ( 63.5^{\circ}\right ) = \overline{CD}$

$\displaystyle 7.16 = \overline{CD}$

The length of the entire chord is twice the length of $\displaystyle \overline{CD}$, so the entire chord length is $\displaystyle 14.3$.

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

$\displaystyle \small sin(27.45)=\frac{x}{47.25}$

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

$\displaystyle \small 47.25*sin(27.45)=x$

$\displaystyle \small x=21.7810391968006$ or $\displaystyle \small 21.78$

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 top of the building. We know that the sine of an angle is equal to the ratio of the side opposite to that angle to the hypotenuse of the triangle. Thus, for our triangle, we know:

$\displaystyle \small sin(25.5) =\frac{30}{x}$

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

$\displaystyle \small \small x=\frac{30}{sin(25.5)}$

This is $\displaystyle \small 69.6846149202954$. Now, take the decimal portion in order to find the number of inches involved.

$\displaystyle \small 0.6846149202954 * 12 = 8.2153790435448$

 Thus, rounded, your answer is $\displaystyle \small 69$ feet and $\displaystyle \small 8$ inches.

To find the sine of an angle, remember the mnemonic SOH-CAH-TOA. 
This means that 

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

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

We are asked to find the $\displaystyle \sin(A)$. So at point $\displaystyle A$ we see that side $\displaystyle a$ is opposite, and the hypotenuse never changes, so it is always $\displaystyle c$. Thus we see that 
$\displaystyle \sin(A) = \frac{a}{c}$

In right triangles, SOHCAHTOA tells us that $\displaystyle \sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}$, 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.

$\displaystyle \sin 42^{\circ} = \frac{CB}{25}$

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

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

Thus, $\displaystyle 17.5= CB$.

In right triangles, SOHCAHTOA tells us that $\displaystyle \sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}$, and we know that $\displaystyle \angle A = 33^{\circ}$ and hypotenuse $\displaystyle AC = 5400$. Therefore, a simple substitution and some algebra gives us our answer.

$\displaystyle \sin 33^{\circ} = \frac{CB}{5400}$

$\displaystyle .54 = \frac{CB}{5400}$ Use a calculator or reference to approximate cosine.

$\displaystyle 2916= CB$ Isolate the variable term. 

Thus, $\displaystyle 2916= CB$.

In right triangles, SOHCAHTOA tells us that $\displaystyle \sin A = \frac{\textup{opposite}}{\textup{hypotenuse}}$, and we know that $\displaystyle \angle A = 45^{\circ}$ and hypotenuse $\displaystyle AC = 17$. Therefore, a simple substitution and some algebra gives us our answer.

$\displaystyle \sin 45^{\circ} = \frac{CB}{17}$

$\displaystyle 17\cdot \sin 45^{\circ}= CB$ Isolate the variable term.

Thus, $\displaystyle 12.02= CB$.