# Thread: law of cosines (distance)

1. ## law of cosines (distance)

a 100 ft verticle tower is to be erected on a side of a hill that makes a 6 degree angle rising up. find the length of each of the guide wires that will be anchored 75 ft uphill and hown hill from the base of the tower

2. So you have two triangles with sides of 100 and 75 feet. The angle between those sides for the uphill cable is 90- 6= 94 degrees. The angle for the downhill cable is 90+ 6= 96 degrees. Now, use the cosine law.

3. Hello, kenzie103109!

A 100-ft vertical tower is to be erected on a side
of a hill that makes a 6° angle with the horizontal.
Find the length of each of the guide wires that will be anchored 75 ft
uphill and downhill from the base of the tower.
Code:
                        D
*
|
* |  *
|
*   |     *  L2
L1      |100
*     |        *
|                 * P
*       |           o
96° | 84° *     B
*         o  96°
A      * 84° |C
o 6°        |
Q * - - - - - - - - * - - - - - - - - * R

The slope of the hill is $PQ\!:\;\angle PQR = 6^o$
The tower is $CD = 100.$

The guide wires are anchord at $A$ and $B.$
. . $AC \,=\,CB\,=\,75.$

The guide wires are: . $L_1 = DA,\;L_2 = DB.$

We find that: . $\angle DCA = 96^o,\;\angle DCB = 84^o.$

In $\Delta DCA\!:\;\;L_1^2 \:=\:25^2 + 100^2 - 2(25)(100)\cos96^o \:=\:11,\!147,64232$

. . Therefore: . $L_1 \;\approx\;105.6\text{ ft}$

In $\Delta DCB\!:\;\;L_2^2 \;=\;25^2 + 100^2 - 2(25)(100)\cos84^o \;=\;10,\!727.35768$

. . Therefore: . $L_2 \;\approx\;103.6\text{ ft}$