# finding length

• Dec 1st 2006, 03:47 AM
zak_killcity14@hotmail.co
finding length
A Steel chimney is held in position by several guy wires. Two such wires AB and AC where B,C and the base of the chimney,D,are on the same level in a staight line with B and C on the left of the chimney. If Angle ABC=45 degrees, ACD=67 degrees and BC=32m. Find the hight of A above D?:confused:
• Dec 1st 2006, 04:52 AM
malaygoel
Quote:

Originally Posted by zak_killcity14@hotmail.co
A Steel chimney is held in position by several guy wires. Two such wires AB and AC where B,C and the base of the chimney,D,are on the same level in a staight line with B and C on the left of the chimney. If Angle ABC=45 degrees, ACD=67 degrees and BC=32m. Find the hight of A above D?:confused:

the figure is like this:
make a right angled triangle, right angled at D
label one vertex as A, other as B
On DB, choose a point C
Now, angleABD(or ABC)=45degrees
and angleACD=67degrees
BC=32 m
from triangle ACD
$\frac{AD}{CD}=tan67$
or $CD=\frac{AD}{tan67}......(1)$
from triangle ABD
$\frac{AD}{BD}=tan45$
or $BD=\frac{AD}{tan45}......(2)$
Subtracting (1) from (2)
$BD-CD=\frac{AD}{tan45} - \frac{AD}{tan67}$
but BD-CD=BC=32
hence
$32=\frac{AD}{tan45} - \frac{AD}{tan67}$
$AD=$ $\frac{32}{\frac{1}{tan45} - \frac{1}{tan67}}$

Keep Smiling
Malay
• Dec 1st 2006, 05:24 AM
Soroban
Hello, Zak!

I assume you have a sketch . . .

Quote:

A steel chimney is held in position by several guy wires.
Two such wires $AB$ and $AC$ where $B,\,C$ and the base of the chimney, $D$,
are on the same level in a straight line with $B$ and $C$ on the left of the chimney.
If $\text{angle }ABC = 45^o,\text{ angle }ACD = 67^o,\;BC = 32\text{ m,}$
find the height of the chimney, $AD.$

Code:

                              * A                           * *|                         *  * |                     *    *  |                   *      *  | y               *        *    |             *          *    |         * 45°        * 67°  |       *---------------*-------*       B      32      C  x  D

Let $x = CD,\:y = AD.$

In right triangle $ADB:\;\tan45^o = \frac{y}{x+32}\quad\Rightarrow\quad1 = \frac{y}{x+32}\quad\Rightarrow\quad x \:= \:y - 32$ [1]

In right triangle $ADC:\;\tan67^o = \frac{y}{x}\quad\Rightarrow\quad x \,=\,\frac{y}{\tan67^o}$ [2]

Equate [1] and [2]: . $y - 32 \:=\:\frac{y}{\tan67^o}$

Solve for $y:\;\;y\tan67^o - 32\tan67^o \:=\:y\quad\Rightarrow\quad y\tan67^o - 32\tan67^o \:=\:y$

. . $y\tan67^o - y \:=\:32\tan67^o\quad \Rightarrow\quad y(\tan67^o - 1) \:=\:32\tan67^o$

Therefore: . $y \;=\;\frac{32\tan67^o}{\tan67^o - 1} \;=\;55.60138965 \;\approx\;\boxed{55.6\text{ m}}$