Need some help ! newbie here !

• Jan 11th 2013, 05:55 PM
jhaycianofficial
Need some help ! newbie here !
hi i'm Jhaycian Velasquez 19 years of age an engineering student :))

two observers who are 2 miles apart on a horizontal plane observe a balloon in a same vertical plane with them

two observers who are 2 miles apart on a horizontal plane observe a balloon in a same vertical plane with themselves. the angles of elevation are 50 degrees and 65 degrees respectively. find the height of the balloon,(a)if it is between the observers;(b)if it is on the same side of both of them.

:)) i hope u will help me :))
• Jan 11th 2013, 07:33 PM
Soroban
Re: Need some help ! newbie here !
Hello, Jhaycian!

Quote:

Two observers are 2 miles apart on a horizontal plane.
They observe a balloon in a same vertical plane with themselves.
The angles of elevation are 50 degrees and 65 degrees respectively.
Find the height of the balloon

(a) if it is between the observers.

Code:

            C             *           *| *           * |  *         *  |h    *         *  |      *       *65  |      50 *     A *-----*-----------* B       :  x  D    2-x    :
The observers are at $A$ and $B$: $AB = 2.$
The balloon is at $C$; its height is $h = CD.$
$\angle A = 65^o,\;\angle B = 50^o.$
Let $x = AD\quad\Rightarrow\quad 2-x = DB.$

In $\Delta CDA\!:\;\tan65^o \,=\,\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan65^o}$ .[1]

In $\Delta CDB\!:\;\tan50^i \:=\:\frac{h}{2-x} \quad\Rightarrow\quad x \:=\:\frac{2\tan50^o - h}{\tan50^o}$ .[2]

Equate [1] and [2]: . $\frac{h}{\tan65^o} \:=\:\frac{2\tan50^i - h}{\tan50^o}$

. . . . . . . . . . . . . . $h\tan50^o \:=\:2\tan50^o\tan65^o - h\tan65^o$

. . . . . . . $\h\tan50^o + h\tan65^o \;=\;2\tan50^o\tan65^o$

. . . . . . $h(\tan50^o + \tan65^o) \;=\;2\tan50^o\tan65^o$

Therefore: . $h \;=\;\frac{2\tan50^o\tan65^o}{\tan50^o+\tan65^o} \;\;\approx\;\;1.532\text{ miles.}$

Quote:

(b) if it is on the same side of both of them.

This is the diagram for this scenario.
Code:

                      * C                     **|                   * * |                 *  *  |               *  *  |h             *    *    |           *    *    |         *50    *65    |     A *-------*-------* D       :  2  B  x  :
Give it a try . . .
• Jan 12th 2013, 04:02 AM
jhaycianofficial
Re: Need some help ! newbie here !
Quote:

Originally Posted by Soroban
Hello, Jhaycian!

Code:

            C             *           *| *           * |  *         *  |h    *         *  |      *       *65  |      50 *     A *-----*-----------* B       :  x  D    2-x    :
The observers are at $A$ and $B$: $AB = 2.$
The balloon is at $C$; its height is $h = CD.$
$\angle A = 65^o,\;\angle B = 50^o.$
Let $x = AD\quad\Rightarrow\quad 2-x = DB.$

In $\Delta CDA\!:\;\tan65^o \,=\,\frac{h}{x} \quad\Rightarrow\quad x \:=\:\frac{h}{\tan65^o}$ .[1]

In $\Delta CDB\!:\;\tan50^i \:=\:\frac{h}{2-x} \quad\Rightarrow\quad x \:=\:\frac{2\tan50^o - h}{\tan50^o}$ .[2]

Equate [1] and [2]: . $\frac{h}{\tan65^o} \:=\:\frac{2\tan50^i - h}{\tan50^o}$

. . . . . . . . . . . . . . $h\tan50^o \:=\:2\tan50^o\tan65^o - h\tan65^o$

. . . . . . . $\h\tan50^o + h\tan65^o \;=\;2\tan50^o\tan65^o$

. . . . . . $h(\tan50^o + \tan65^o) \;=\;2\tan50^o\tan65^o$

Therefore: . $h \;=\;\frac{2\tan50^o\tan65^o}{\tan50^o+\tan65^o} \;\;\approx\;\;1.532\text{ miles.}$

This is the diagram for this scenario.
Code:

                      * C                     **|                   * * |                 *  *  |               *  *  |h             *    *    |           *    *    |         *50    *65    |     A *-------*-------* D       :  2  B  x  :
Give it a try . . .

Thanks a lot for help me out :D