Since $△ABC$ is right at $C$ and $CD$ is an altitude to the hypotenuse $AB$, the altitude theorem applies. First, find the hypotenuse: $AB=AD+DB=9+16=25$. Then

so $CD=12$. In the two smaller right triangles, angle $A$ is the same as in $△ACD$, so

Similarly, angle $B$ is the same as in $△BCD$, so

Therefore,

So the correct answer is $\frac{25}{12}$.

Let $x$ be the horizontal distance from point $B$ to the point on the ground directly below $D$, and let $h$ be the height of $D$ above the ground.

From point $B$, the angle of elevation is $45^{\circ }$, so

which means

Point $A$ is 120 meters farther from the point directly below $D$ than point $B$, so the horizontal distance from $A$ is $x+120$.

From point $A$, the angle of elevation is $30^{\circ }$, so

Substitute $h=x$:

Multiply both sides by $\sqrt{3}(x+120)$:

Now solve:

Rationalize the denominator:

Since $h=x$, the height is

Since $\tan \theta =\frac{3}{4}$, the ratio of the legs opposite and adjacent to $\theta$ is $3:4$. So the legs can be written as $3k$ and $4k$ for some positive constant $k$. The area of a right triangle is $\frac{1}{2}(\text{leg})(\text{leg})$, so

This simplifies to

so

and therefore $k=2$. The legs are $6$ and $8$. A triangle with side ratio $3:4:5$ has hypotenuse $5k$, so the hypotenuse is

Therefore, the statement that must be true is that the hypotenuse is $10$.

Let the acute angle opposite the shorter leg be $\theta$. Since the shorter leg is opposite $\theta$ and the longer leg is adjacent to $\theta$,

Cross-multiply:

So

which gives

Then the longer leg is

The hypotenuse is found with the Pythagorean theorem:

Therefore, the hypotenuse is $35$ inches.

Let the horizontal distance from the closer observation point to the building be $x$ feet. Since the surveyor walked 80 feet toward the building, the horizontal distance from the farther point is $x+80$ feet. Let the building's height be $h$ feet.

Using the closer point, where the angle of elevation is $45^{\circ }$:

so

Using the farther point, where the angle of elevation is $30^{\circ }$:

Substitute $h=x$:

Multiply both sides by $\sqrt{3}(x+80)$:

Rearrange:

So

Rationalize the denominator:

Because $h=x$, the height of the building is