Answer :

Given:

Let AB and CD be the cross bars of the football goal.

Let O be a point, 'a' metres behind and opposite the middle of the football goal.

Let 'l' metres be the length of the field.

Let AB = CD = p m since AB and CD are the cross bars of the football goal.

In right angled ΔBAO,

AB / AO = tan α

AO = p / tan α

In right angled ΔDCO,

CD / CO = tan β

CO = p / tan β

Length of the field = AO + OC

= a + (l - a)

= ( p / tan α) + ( p / tan β)

By replacing p = a tan α we get,

= a (1 + tan α cotβ) m

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