The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m, find the height of the tower from the base of the tower and in the same straight line with it are complementary.
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Let the height of the tower = h m
<B = 90°
C and D are the two points such that
BC = 4 m and BD = 9 m
<BCA = x°
<BDA = ( 90 - x )°
i ) In ∆ABC ,
tan x° = AB/BC
tan x° = h/4 ------( 1 )
ii ) In ∆ABD ,
tan( 90 - x )° = AB/BD
cot x° = h/9 -----( 2 )
multiply ( 1 ) and ( 2 ) , we get
( tan x° )(cot x° ) = ( h/4 )( h/9 )
[ We know that , tanx ° cotx° = 1 ]
1 = h²/36
h² = 36
h = √36
h = 6 m
Therefore ,
Height of the tower = h = 6 m
I hope this helps you.
: )
<B = 90°
C and D are the two points such that
BC = 4 m and BD = 9 m
<BCA = x°
<BDA = ( 90 - x )°
i ) In ∆ABC ,
tan x° = AB/BC
tan x° = h/4 ------( 1 )
ii ) In ∆ABD ,
tan( 90 - x )° = AB/BD
cot x° = h/9 -----( 2 )
multiply ( 1 ) and ( 2 ) , we get
( tan x° )(cot x° ) = ( h/4 )( h/9 )
[ We know that , tanx ° cotx° = 1 ]
1 = h²/36
h² = 36
h = √36
h = 6 m
Therefore ,
Height of the tower = h = 6 m
I hope this helps you.
: )
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