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Question
Two Navy helicopters A and B are flying over the Bay of Bengal at saine altitude from sea level to search a missing boat. Pilots of both the helicopters sight the boat at the same time while they are apart 10 km from each other. If the distance of the boat from A is 6 km and if the line segment AB subtends 60° at the boat, find the distance of the boat from B
Solution
A, B are the positions of the helicopter above the sea level.
Distance between A and B = 10 km
C – Position of the boat on the surface of sea.
AC, BC are the directions of the boat as seen from A and B respectively.
Distance of the boat C from A = 6 k.m
∠ACB = 60°
Using cosine formula
AB2 = BC2 + AC2 – 2 BC . AC cos ∠ACB
c2 = a2 + b2 – 2ab cos C
102 = a2 + 62 – 2a × 6 cos 60°
100 = `"a"^2 + 36 - 12"a"(1/2)`
0 = a2 + 36 – 6a – 100
a2 – 6a – 64 = 0
a = `(6 +- sqrt(36 - 4(1)(- 64)))/(2 xx 1)`
= `(6 +- sqrt(36 + 256))/2`
= `(6 +- sqrt(292))/2`
= `(6 +- sqrt(4 xx 73))/2`
a = `(6 +- 2sqrt(73))/2`
= `3 +- sqrt(73)`
a = `3 + sqrt(73)` or a = `3 - sqrt(73)`
a = `3 - sqrt(73)` is not possible.
∴ a = `3 + sqrt(73)` km
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