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प्रश्न
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?
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उत्तर
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.
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