A hunter spots a bird flying in the sky at an elevation 60 $^{\circ}$ from his line of sight. Keeping in mind the speed of the bird, he fires his gun at an elevation 45 $^{\circ}$ with his line of sight. However, the bird changes its direction by the time the bullet crosses the bird, the bird has flown down to an elevation 30º from his line of sight. At the time the bullet and the bird are vertically in-line, what would be the distance between them (Assume initial height of the bird from the ground = 2000m)?

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Solution

The correct option is A

Given that: AY= BX = 2000m
Also, $∠$ APX= 60 $^{\circ}$ , $∠$ BPX= 45 $^{\circ}$ and $∠$ CPX = 30 $^{\circ}$ .
PX= 2000m (Since PBX is an isosceles right-angled triangle).
Also, CXP is a 30-60-90 triangle $\Rightarrow$ CX = $\frac{1000}{\sqrt{3}}$ m
Therefore, required distance BC =
2000 - $\frac{1000}{\sqrt{3}}$ m

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The correct option is A Given that: AY= BX = 2000m Also, ∠APX= 60∘, ∠BPX= 45∘ and ∠CPX = 30∘. PX= 2000m (Since PBX is an isosceles right-angled triangle). Also, CXP is a 30-60-90 triangle ⇒ CX = 1000√3m Therefore, required distance BC = 2000 - 1000√3m