The angles of elevation of a hot air balloon
from two points, A and B, on level ground,
are 24.2° and 46.8°, 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.
Answers
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 = HOC=H
From triangle AOC, using cotangent of angle 'a', we have:
\begin{gathered}\cot a = \frac{AC}{OC}\\\\AC=H\cot a------(1)\end{gathered}
Step-by-step explanation:
AC=Hcota−−−−−−(1)
From triangle BOC, using cotangent of angle 'b', we have:
\begin{gathered}\cot b = \frac{BC}{OC}\\\\BC=H\cot b------(2)\end{gathered}
NoBC
BC=Hcotb−−−−−−(2)
Now, AB = AC + BCAB=AC+BC
\begin{gathered}H\cot a + H\cot b = 8.4\\\\H(\cot a +\cot b) = 8.4\end{gathered}
Hcota+Hcotb=8.4
H(cota+cotb)=8.4
Plug in the values of 'a' and 'b' and solve for 'H'. This gives,
\begin{gathered}H=\dfrac{8.4}{\cot(24.2)+\cot(46.8)}\\\\H=2.65\ miles\end{gathered}
H=
cot(24.2)+cot(46.8)
8.4
H=2.65 miles