Question

The angles of elevation of a hot air balloon from two points, A and B, on level ground, are 24.2 and 46.8 degrees, respectively. The points A and B are 8.4 miles apart, and the balloon is between the points in the same vertical plane. Find the height of the balloon above the ground.

Answer 1

Answer:

The height of the balloon above the ground is 2.65 miles.

Step-by-step explanation:

Given:

Angle of elevation of balloon from point A (a) = 24.2°

Angle of elevation of balloon from point B (b) = 46.8°

The distance between A and B (AB) = 8.4 miles

Consider two right angled triangles representing the given scenario.

Let the height of balloon be 'H'. So, $OC = H$

From triangle AOC, using cotangent of angle 'a', we have:

$\cot a = \frac{AC}{OC}\\\\AC=H\cot a------(1)$

From triangle BOC, using cotangent of angle 'b', we have:

$\cot b = \frac{BC}{OC}\\\\BC=H\cot b------(2)$

Now, $AB = AC + BC$

$H\cot a + H\cot b = 8.4\\\\H(\cot a +\cot b) = 8.4$

Plug in the values of 'a' and 'b' and solve for 'H'. This gives,

$H=\dfrac{8.4}{\cot(24.2)+\cot(46.8)}\\\\H=2.65\ miles$

Therefore, the height of the balloon above the ground is 2.65 miles.