At what time between 6 and 7 will the hands be perpendicular "explanation"
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Hi ,
*******************************************
1 )The clock is divided into 60 equal
minute divisions .
2 ) One minute division = 360°/60° = 6 a part
3 ) In one minute , the hour moves 1°/2
4 ) In one minute , the minute hand gains
5½° more than hour hand .
*******************************************
H = Initial position of hour hand
M = minutes
x = angle
Here ,
hands are perpendicular to each other
=> x = 90°
H = 6
If x = 90° then the hands of clock are
right angles twice in every hours .
i ) 30H - 11/2( M ) = x°
30× 6 - 11/2 ( M ) = 90°
180 - 90 = 11/2 ( M )
( 90 × 2 )/11 = M
180/11 = M
M = 16 4/11
=> 16 4/11 minutes after 6' O clock
ii ) 30( H ) - 11/2 ( M ) = - x°
30× 6 - 11/2 ( M ) = - 90°
180 + 90 = 11/2( M )
( 270 × 2 )/11 = M
540/11 = M
M = 49 1/11
Therefore ,
=> 49 1/11 minutes after 6' O clock
I hope this helps you.
: )
*******************************************
1 )The clock is divided into 60 equal
minute divisions .
2 ) One minute division = 360°/60° = 6 a part
3 ) In one minute , the hour moves 1°/2
4 ) In one minute , the minute hand gains
5½° more than hour hand .
*******************************************
H = Initial position of hour hand
M = minutes
x = angle
Here ,
hands are perpendicular to each other
=> x = 90°
H = 6
If x = 90° then the hands of clock are
right angles twice in every hours .
i ) 30H - 11/2( M ) = x°
30× 6 - 11/2 ( M ) = 90°
180 - 90 = 11/2 ( M )
( 90 × 2 )/11 = M
180/11 = M
M = 16 4/11
=> 16 4/11 minutes after 6' O clock
ii ) 30( H ) - 11/2 ( M ) = - x°
30× 6 - 11/2 ( M ) = - 90°
180 + 90 = 11/2( M )
( 270 × 2 )/11 = M
540/11 = M
M = 49 1/11
Therefore ,
=> 49 1/11 minutes after 6' O clock
I hope this helps you.
: )
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