Question

Q. A round balloon of radius _r_ subtends an angle _a_ at the eye of the observer while the angle of elevation of its centre is _b_ . Prove that height of the centre of the balloon is _r_ sin _b_ cosec _a_/2.
Please also draw the figure of the given situation.

Answer 1

Hello,let's look at the figure.
Let the height of centre of the balloon above the ground be h m.
Given, balloon subtends an θ angle at the observes eye.
∠ EAD = θ
In ΔACE and ΔACD:
AE = AD (length of tangents drawn from an external point to the circle are equal) ;
AC = AC (common);
 CE = CD (radius of the circle)

ΔACE ≈ ΔACD  (SSS congruence criterion)
⇒ ∠EAC = ∠DAC  (CPCT)
EAC = DAC = θ/2

In right ΔACD,
sin θ/2= CD/AC;
sin θ/2= a/AC
⇒AC=a/sin θ/2= a cosecθ/2         (1)

In right ΔACB,
sinΦ=BC/AC;
using the (1)
sinΦ=h/ a cosecθ/2 ;
⇒ h=a sinΦ cosecθ/2

Thus, the height of the centre of the balloon is:
a·sinΦ· cosecθ/2

bye :-)