# Thread: word problem

1. ## word problem

Vivian and Bobby are 250m apart and are facing each other. Each one is
looking up at a hot air balloon. The angle of elevation from Vivian to the
balloon is 75° and the angle of elevation from Bobby to the balloon is 50°.
Determine the height of the balloon, to one decimal place.

2. let Viv be a horizontal distance x to a point directly beneath the balloon.
Bob is (250-x) from the same point.

extend a perpendicular vertically downward from the balloon to the ground ... let that vertical distance = h.

note also that two right triangles are formed ... one for Viv, one for Bob.

using Viv's triangle, h/x = tan(75)

using Bob's triangle, h/(250-x) = tan(50)

you now have two equations with two unknowns ... your goal is to find the value of h.

3. Hello, euclid2!

I like skeeter's solution . . . Here's my version.

Vivian and Bobby are 250 m apart and are facing each other.
Each one is looking up at a hot air balloon.
The angle of elevation from Vivian to the balloon is 75°
and the angle of elevation from Bobby to the balloon is 50°.
Determine the height of the balloon, to one decimal place.
Code:
              H
*
*: *
* 55° *
*  :     *  v
*   :       *
*    :y        *
*     :           *
* 75°  :        50°  *
V *   *   *   *   *   *   * B
h = 250

We have: .$\displaystyle VB = h = 250$
Since $\displaystyle \angle V = 75^o,\;\angle B = 50^o$, we have: .$\displaystyle \angle H = 55^o$
And we want $\displaystyle y.$

Law of Sines: .$\displaystyle \frac{v}{\sin V} = \frac{h}{\sin H} \quad\Rightarrow\quad v \:=\:\frac{250\sin75^o}{\sin55^o} \:\approx\:294.8$

We have: .$\displaystyle \sin B = \frac{y}{v}\quad\Rightarrow\quad y \:=\:v\!\cdot\!\sin B \:=\:294.8\sin50^o \;=\;\boxed{225.8\text{ m}}$

4. Thanks very much