Hello, qbkr21!

A hot-air balloon is floating above a straight road.

To calculate their height above the ground, the balloonists simultaneously

measure the angle of depression to two consecutive mileposts

on the road on the same side of the balloon.

The angles of depression are found to be 23° and 25°.

How high (in feet) is the ballon?

The observation to the nearer milepost looks like this: Code:

B * - - - - - - - - - - - - - - -
: o 25°
: o
: o
: o
: o
: o
: 25° o
--+---------------o---------------*--
: - - - 1 - - - :

The observation to the further milepost looks like this: Code:

B * - - - - - - - - - - - - - - -
: * 23°
: *
: *
: *
: *
: *
: 23° *
--+---------------o---------------*--
: - - - 1 - - - :

Combining the diagrams we have: Code:

B *
: o *
: o *
: o *
y : o *
: o *
: o *
: 25° o 23° *
--+---------------o---------------*--
: - - - x - - - : - - - 1 - - - :

We have: .$\displaystyle \tan25^o \,=\,\frac{y}{x}\quad\Rightarrow\quad y \,=\,x\!\cdot\!\tan25^o$ **[1]**

And: .$\displaystyle \tan23^o \,= \,\frac{y}{x+1}\quad\Rightarrow\quad y \,=\,(x+1)\tan23^o $ **[2]**

Equate [1] and [2]: .$\displaystyle x\tan25^o \:=\:(x + 1)\tan23^o $

. . . . . . . . . . . . . . .$\displaystyle x\tan25^o \:=\:x\tan23^o + \tan23^o$

. . . . . . . $\displaystyle x\tan25^o - x\tan23^o \:=\:\tan23^o$

. . . . . . . $\displaystyle x\left(\tan25^o - \tan23^o\right) \:=\:\tan23^o$

. . . . . . . . . . . . . . . . . . $\displaystyle x \:=\:\frac{\tan23^o}{\tan25^o - \tan23^o} \:\approx\:10.147$

Substitute into [1]: .$\displaystyle y \:=\:(10.147)\tan25^o \:=\:4.73162..$ miles

Therefore, the height of the balloon is about $\displaystyle \boxed{4.7\text{ miles.}}$