The angle of elevation of an object from a point P on the level g

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321.

The angle of elevation of an object from a point P on the level ground is α. Moving d metres on the ground towards the object, the angle of elevation is found to be β. Then the height (in metres) of the object is

  • dtanα

  • dcotβ

  • dcotα + cotβ

  • dcotα - cotβ


D.

dcotα - cotβ

Let the height of the object is h.

In ABC,      tanα = hx + d x +d = hcotβ   . . .iAnd in   ABD,tanβ = hx   x = hcotβOn putting this value in Eqs. i, we get                 hcotβ +d = hcotα hcotα - cotβ = d h = dcotα - cotβ


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322.

A vertical pole subtends an angle tan-112 at a point P on the ground. If the angles substended by the upper half and the lower half of the pole at P are respectively α and β , then tanα, tanβ is equal to

  • 14, 15

  • 15, 29

  • 29, 14

  • 14, 29


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