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Question
Two soldiers A and B in two different underground bunkers on a straight road, spot an intruder at the top of a hill. The angle of elevation of the intruder from A and B to the ground level in the eastern direction are 30° and 45° respectively. If A and B stand 5km apart, find the distance of the intruder from B
Solution
Let A and B be the two positions of the soldiers.
AC – direction of the intruder seen from A.
BC – the direction of the intruder seen from B.
∠BAC = 30° angle of elevation of the intruder from A.
∠PBC = 45° angle of elevation of the intruder from B.
Distance between A and B = 5k.m.
In ∆ABC,
∠ABC = 180° – 45° = 135°
∠BCA = 180° – (135° + 30°)
= 180° – 165°
= 15°
Using sine formula
`"BC"/(sin∠"BAC") = "AB"/(sin∠"ACB")`
`"BC"/(sin 30^circ) = 5/(sin 15^circ)`
`"BC"/(1/2) = 5/(sin(45^circ - 30^circ))`
2BC = `5/(sin 45^circ cos 30^circ - cos 45^circ sin 30^circ)`
2BC = `5/(1/sqrt(2) * sqrt(3)/2 - 1/sqrt(2) * 1/2)`
2BC = `5/((sqrt(3) - 1)/(2sqrt(2))`
= `(2sqrt(2) xx 5)/(sqrt(3) - 1)`
BC = `(5sqrt(2))/(sqrt(3) - 1)` k.m.
Distance of intruder from B = `(5sqrt(2))/(sqrt(3) - 1)` k.m.
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