Let the height of the balloons be x meters.
In triangle AOX,
tan(14°15') = x/1000
In triangle BOX,
tan(10°30') = x/1000
Now, solve for x in both equations:
tan(14°15') = x/1000
x = 1000 * tan(14°15')
x ≈ 243.89 meters
Therefore, the height of the balloons is approximately 243.89 meters.
A and B are two balloons at the same height and 1000m apart. O is ground observation position,angle AOX =14°15' and angle BOX=10°30'.Find the height of the balloons
3 answers
where is X?
On the ground? On the line AB?
On the ground? On the line AB?
I apologize for the confusion. Let's correct the solution.
Let's consider the triangle AOB with side lengths as 1000m (AB), height of the balloons (h), and the distance of the observer position O from line AB. Let's say the distance of O from line AB is d meters.
In triangle AOX, tan(14°15') = h/d
In triangle BOX, tan(10°30') = h/(1000 - d)
We know that the observer is ground observation positioned at O. So, the distance of O from line AB is 0 (d=0).
Let's substitute the values in the equation:
tan(14°15') = h/0 => h=0
This means that the height of the balloons is 0 meters, and the observer is directly below the line joining the two balloons.
Let's consider the triangle AOB with side lengths as 1000m (AB), height of the balloons (h), and the distance of the observer position O from line AB. Let's say the distance of O from line AB is d meters.
In triangle AOX, tan(14°15') = h/d
In triangle BOX, tan(10°30') = h/(1000 - d)
We know that the observer is ground observation positioned at O. So, the distance of O from line AB is 0 (d=0).
Let's substitute the values in the equation:
tan(14°15') = h/0 => h=0
This means that the height of the balloons is 0 meters, and the observer is directly below the line joining the two balloons.