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A Man on the Top of a Tower Observes a Truck at an Angle of Depression ∝ Where ∝ = 1 √ 5 and Sees that It is Moving Towards the Base of the Tower. Ten Minutes Later, the Angle of Depre - Mathematics

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प्रश्न

A man on the top of a tower observes a truck at an angle of depression ∝ where `∝ = 1/sqrt(5)` and sees that it is moving towards the base of the tower.  Ten minutes later, the angle of depression of the truck is found to `β = sqrt(5)`. Assuming that the truck moves at a uniform speed, determine how much more ti me it will take to each the base of the tower? 

योग

उत्तर

In the figure , AB is the tower . A is the position of the man. C and D are the two positions of the truck.

Let the speed of the truck be x m/sec

Distance CD = Speed × time = 600x

In right triangle ABC,

`tan α = "h"/"BC"`

It is a given that `tan α = 1/sqrt(5)`

`"BC" = "h"sqrt(5)`   .. (1)

In right triangle ABD,

`tan β = "h"/"BD"`

It is given that `tan β = sqrt(5)`

`"h" = sqrt(5)"BD"`

Now , CD = BC - BD

600x = 5BD - BD

BD = 150x

Time taken = `(150"x")/"x"` = 150 seconds

Thus , the time taken by the truck to reach the tower is 150 sec = 2 min 30 sec.

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Heights and Distances - Solving 2-D Problems Involving Angles of Elevation and Depression Using Trigonometric Tables
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 22: Heights and Distances - Exercise

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फ्रैंक Mathematics - Part 2 [English] Class 10 ICSE
अध्याय 22 Heights and Distances
Exercise | Q 47

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