Question

An aeroplane is observed at the same time by two anti aircraft batteries 6000 m apart to be at elevation of 30 degree and 45 respectively. Assuming that the aeroplane is travelling directly towards the two batteries find its height and its horizontal distance from the nearer battery

Answer 1

here is the fully explained above

Answer 2

Answer:

8196.1524 m

Step-by-step explanation:

Refer the attached figure

Let the height of tower be h

Distance of 1 st battery from the base of tower be x

So, Distance of base form 2nd battery = x+600

In ΔABC

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 45^{\circ} = \frac{AB}{BC}$

$1= \frac{AB}{BC}$

$AB= BC$

$h=x$ ---A

In ΔABD

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 30^{\circ} = \frac{AB}{BD}$

$\frac{1}{\sqrt{3}}= \frac{h}{x+6000}$

From A

$\frac{1}{\sqrt{3}}= \frac{x}{x+6000}$

$x+6000=\sqrt{3}x$

$6000=\sqrt{3}x-x$

$6000=(\sqrt{3}-1)x$

$\frac{6000}{(\sqrt{3}-1)}=x$

$8196.1524=x$

Since h = x

So,its height and its horizontal distance from the nearer battery is 8196.1524 m