Question

a ballon of radius r subsrend an angle alpha at the eye of observer while the angle of elevation of the centre ( balloon) is belta. prove that the height of the centre of the balloon is r. sin belta. cos alpha by 2​

Answer 1

Question

A balloon of radius r subsrend an angle alpha at the eye of observer while the angle of elevation of the centre ( balloon) is beta. Prove that the height of the centre of the balloon is r. sin beta. cos (alpha/2)

$\large\underline{\sf{Solution-}}$

Let assume that

• O be the center of ths balloon of radius r

• P be the eye of observer.

Now, PA and PB be tangent lines from eye of the observer at P to the balloon such that

$\rm :\longmapsto\:\angle APB = \alpha$

We know, Line segment joining the center and external point from where tangents are drawn bisects the angle between the tangents.

So,

$\rm\implies \:\angle APO = \angle OPB = \dfrac{ \alpha }{2}$

Now,

From center of the balloon O, draw OC perpendicular to the eye level of the observer.

According to statement, it is given that angle of elevation of the center of the balloon from eye of the observer is

$\rm :\longmapsto\:\angle OPC = \beta$

Now, In right angle triangle OAB

$\rm :\longmapsto\:cosec\dfrac{ \alpha }{2} = \dfrac{OP}{OB}$

$\rm :\longmapsto\:cosec\dfrac{ \alpha }{2} = \dfrac{OP}{r}$

$\rm\implies \:\boxed{\tt{ \: OP \: = \: r \: cosec \: \frac{ \alpha }{2} \: }}$

Now, In right triangle OPC

$\rm :\longmapsto\:sin \beta = \dfrac{OC}{OP}$

$\rm\implies \:OC = OP \: sin \beta$

$\rm\implies \:OC = r \: cosec \dfrac{ \alpha }{2} \: sin \beta$

Hence, Proved

Answer 2

Question

A balloon of radius r subsrend an angle alpha at the eye of observer while the angle of elevation of the centre ( balloon) is beta. Prove that the height of the centre of the balloon is r. sin beta. cos (alpha/2)

$\large\underline{\sf{Solution-}}$

Let assume that

O be the center of ths balloon of radius r

P be the eye of observer.

Now, PA and PB be tangent lines from eye of the observer at P to the balloon such that

$\rm :\longmapsto\:\angle APB = \alpha$

We know, Line segment joining the center and external point from where tangents are drawn bisects the angle between the tangents.

So,

$\rm\implies \:\angle APO = \angle OPB = \dfrac{ \alpha }{2}$

Now,

From center of the balloon O, draw OC perpendicular to the eye level of the observer.

According to statement, it is given that angle of elevation of the center of the balloon from eye of the observer is

$\rm :\longmapsto\:\angle OPC = \beta$

Now, In right angle triangle OAB

$\rm :\longmapsto\:cosec\dfrac{ \alpha }{2} = \dfrac{OP}{OB}$

$\rm :\longmapsto\:cosec\dfrac{ \alpha }{2} = \dfrac{OP}{r}$

$\rm\implies \:\boxed{\tt{ \: OP \: = \: r \: cosec \: \frac{ \alpha }{2} \: }}$

Now, In right triangle OPC

$\rm :\longmapsto\:sin \beta = \dfrac{OC}{OP}$

$\rm\implies \:OC = OP \: sin \beta$

$\rm\implies \:OC = r \: cosec \dfrac{ \alpha }{2} \: sin \beta$

Hence, Proved