# Thread: Need some help ! newbie here !

1. ## 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 )

2. ## Re: Need some help ! newbie here !

Hello, Jhaycian!

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

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

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

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

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

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

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

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

(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 . . .

3. ## Re: Need some help ! newbie here !

Originally Posted by Soroban
Hello, Jhaycian!

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

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

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

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

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

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

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

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

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