Hello, mikel03!

The width of the river between $\displaystyle A$ and $\displaystyle B$ is to be measured.

It is known that $\displaystyle \angle ABD = 90^o,\;\angle ADB = 38^o,\;\angle ACB = 28^o,\;CD = 40$ meters.

Find the width $\displaystyle (w)$. Code:

A *
| * *
| * *
w | * *
| * *
| 38° * 28° *
* - - - - - * - - - - - *
B x D 40 C

We have: .$\displaystyle AB = w,\;\angle ADB = 38^o,\;\angle ACB = 28^o, CD = 40$

Let $\displaystyle BD = x.$

In right triangle $\displaystyle ABD\!:\;\;\tan38^o \:=\:\frac{w}{x} \quad\Rightarrow\quad x \:=\:\frac{w}{\tan37^o}$ .[1]

In right triangle $\displaystyle ABC\!:\;\;\tan28^o \:=\:\frac{w}{x+40}\quad\Rightarrow\quad x \:=\:\frac{w}{\tan28^o} - 40$ .[2]

Equate [2] and [1]: .$\displaystyle \frac{w}{\tan28^o} - 40 \:=\:\frac{w}{\tan38^o} \quad\Rightarrow\quad \frac{w}{\tan28^o} - \frac{w}{\tan38^o} \:=\:40$

Multiply by $\displaystyle \tan28^o\tan38^o\!:\;\;w\tan38^o - w\tan28^o \:=\:40\tan28^o\tan38^o $

Factor: .$\displaystyle w(\tan38^o - \tan28^o) \:=\:40\tan28^o\tan38^o \quad\Rightarrow\quad w \:=\:\frac{40\tan28^o\tan38^o}{\tan38^o-\tan28^o}$

Therefore: .$\displaystyle w \;=\;66.5795769 \;\approx\;66.6$ m.