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A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of `30o` , which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depres

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Let AB be the height of the tower and C,D be the two positions of the car.
`angleDBC=angleDBX-angleCBX=60^(@)-30^(@)=30^(@)`
`:." In "DeltaBDC`,
`angleDBC=angleDCB (30^(@)" each ")`
`angleCD=BD`
(`:.` sides opposite to equal angles are equal)
In right `Delta BDC`,
`cos60^(@)=(AD)/(DB) rArr1/2=(AD)/(DB) rArr DB=2 AD`
Now, time taken to cover the distance CD = 6 sec.
`:. ` Time taken to cover the distance `2AD=6 sec`.
`rArr` Time taken to cover the distance AD = 3 sec.
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