# Help Needed

• Apr 26th 2009, 05:18 PM
mikel03
Help Needed
Hey im having trouble finishing the following question:

The Width of the river between A and B is to be measured. It is known that [ABD = 90 degrees and survey shows that [ADB = 38 degrees [ACB = 28 degrees and distance between C and D is 40 meters. Find out the Width (w)
• Apr 26th 2009, 10:57 PM
aidan
Quote:

Originally Posted by mikel03
Hey im having trouble finishing the following question:
The Width of the river between A and B is to be measured. It is known that [ABD = 90 degrees and survey shows that [ADB = 38 degrees [ACB = 28 degrees and distance between C and D is 40 meters. Find out the Width (w)

As stated there are some assumptions:
1) points C and A are on the opposite side of the river from where points B and D are.
2) the angle BAC is 90 degrees.
3) the banks or edges of the river are parallel.

1) if you are at point B looking at point A, then you need to know if D is 90 degrees to your left or 90 degrees to you right.
2) if you are at point A looking at point B, then you also need to know if C is 90 degrees to your left or 90 degrees to you right.
You have 4 possible solutions.
No wonder you are having trouble with it.
• Apr 27th 2009, 12:53 AM
mikel03
Thanks Aidan,

Ive attached the question, thanks
• Apr 27th 2009, 02:57 AM
Soroban
Hello, mikel03!

Quote:

The width of the river between $A$ and $B$ is to be measured.
It is known that $\angle ABD = 90^o,\;\angle ADB = 38^o,\;\angle ACB = 28^o,\;CD = 40$ meters.
Find the width $(w)$.

Code:

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

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

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

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

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

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

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

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

• Apr 27th 2009, 04:56 AM
Trigonometry
Hello mikel03
Quote:

Originally Posted by mikel03
Hey im having trouble finishing the following question:

The Width of the river between A and B is to be measured. It is known that [ABD = 90 degrees and survey shows that [ADB = 38 degrees [ACB = 28 degrees and distance between C and D is 40 meters. Find out the Width (w)

... or, if you want the easy method:

$\angle DAC = 10^o$ (Exterior angle of triangle = sum of interior opposite angles)

$\Rightarrow \frac{AD}{\sin 28^o}= \frac{40}{\sin 10^o}$ (Sine Rule on $\triangle ADC$)

$\Rightarrow AB = AD \sin 38^o = \frac{40\sin 38^o\sin 28^o}{\sin 10^o} \approx 66.6$