Question

A round balloon of radius '$a$ subtends an angle $\theta$ at the eye of the observer while angle of elevation of its centre is $\varphi$.Then the height of the centre of the balloon is $a\sin \varphi \text{cosec}\frac{\alpha }{2}$.

• A. True
• B. False

Answer 1

The correct option is A True

let the height of the center of the baloon above the ground be $hm$

given- baloon subtends an angle $\alpha$ at the observer's eye

therefore $\angle EAD=\theta$

in $△ACE$ and $△ACD$

$AE=AD$ (length of the tangent drawn from external point to the circle are equal)

$AC=AC$ (common)

$CE=CD$ (radius of the circle)

$△ACE\cong △ACD$ (sss congurence criteria)

$\angle EAC=\angle DAC$ (cpct)

$\angle EAC=\angle DAC=\alpha /2$

$△ACD$ ,

$\sin \frac{\alpha }{2}=\frac{CD}{AC}$

$\sin \frac{\alpha }{2}=\frac{a}{AC}$

$AC=\frac{CD}{\sin \alpha /2}=a\csc \alpha /2$

$△ACB$

$\sin \phi =\frac{BC}{AC}\sin \phi =\frac{h}{a\csc \alpha /2}h=a\csc \alpha /2\sin \phi$

thus, height of the centre of the balloon is $h=a\csc \alpha /2\sin \phi$