Question

A fighter jet has to hit a small target by flying a horizontal distance. When the target is sighted, the pilot measures the angle of depression to be 30°. If after 100 km, the target has an angle of depression of 45°, how far is the target from the fighter jet at that instant?

Answer 1

Let A be the position of the jet fighter observing the target at an angle of depression 30°.

Also, Let B be the position of the jet 100 k.m away horizontally from A observing the target at an angle of depression 45°.

In ∆TAB,

AB = 100 km

∠TAB = 30°

∠ABT = 180°- 45° = 135°

∠ATB = 180° – (135°+ 300)

= 180° – 165°

= 15°

In ∆ABT, `"BT"/(sin 30^circ) = "AB"/(sin 15^circ)`

BT = `100/(sin 15^circ) xx sin 30^circ`

= `100/(sin(45^circ - 30^circ)) xx 1/2`

= `50/(sin 45^circ cos 30^circ -  cos 45^circ sin 30^circ)`

= `50/(1/sqrt(2) xx sqrt(3)/2 - 1/sqrt(2) xx 1/2)`

= `50/((sqrt(3) - 1)/(2sqrt(2))`

= `(50 xx 2sqrt(2))/(sqrt(3) - 1)`

= `(100sqrt(2))/(sqrt(3) - 1) xx (sqrt(3) + 1)/(sqrt(3) + 1)`

= `(100sqrt(2) (sqrt(3) + 1))/(3 - 1)`

= `(100sqrt(2) (sqrt(3) + 1))/2`

= `50sqrt(2) (sqrt(3) + 1)` k.m.

∴ The distance of the target from the position B = `50sqrt(2) (sqrt(3) + 1)` k.m.