Thread: Trying to determine the height of a weather balloon. I think i have it right

1. Trying to determine the height of a weather balloon. I think i have it right

Two observers 5000m apart, standing at A and B measure the angle of elevation to a weather balloon at H as 47degrees and 71degrees. Find the height of the weather balloon correct to the nearest metre.

Not certain about my answer to this question. This is what i came up with

a = c
Sin A Sin C Sin71= opposite
5000m*Sin47=a Hypotenuse
Sin62 Sin71=X
5000m*Sin47=a 4142sin71
a=4142 X= 3916

Is it right ?

2. Re: Trying to determine the height of a weather balloon. I think i have it right

Its gonna be difficult to read. When posting, it manipulates the order that i had written. Now it just disorganized and confusing. Plus i don't know how formulate the math portion.

3. Re: Trying to determine the height of a weather balloon. I think i have it right

your angles are A = 47, B = 71, C = 62

5000/ sin 62 = b/ sin 71

b = 337.83 = 338 m

sin 47 = x/ 338

x = 338 x 0.73 = 246. 74 = 247 m

4. Re: Trying to determine the height of a weather balloon. I think i have it right

wouldn't it then be 5000/Sin62*sin71

If you're right then i guess have some work to do

Thanks for the help

5. Re: Trying to determine the height of a weather balloon. I think i have it right

looks like you are correct.

law of sines ...

$\displaystyle \frac{a}{\sin(71)} = \frac{5000}{\sin(62)}$

$\displaystyle a = \frac{5000\sin(71)}{\sin(62)}$

$\displaystyle h = a \sin(47) \approx 3916 \, m$

6. Re: Trying to determine the height of a weather balloon. I think i have it right

I knew i was right. I should now be able to hand in my work with some confidence.
Thanks

7. Re: Trying to determine the height of a weather balloon. I think i have it right

I've just realized that my calculator was in radians so my sin 47, sin 62, sin 71 were not accurate. Your were right.