type 1 angle between the hands of clock Detailed Explanation And More Example

Answer: (d)

∴ angle traced by hour hand per minute = $(1/2)^o$

Therefore angle traced by hour hand in 4 h 10 min = $[(4 × 60) + 10] × 1°/2 = 250 × 1°/2 = 125°$

∴ angle traced by minute hand per minute = 6°

∴ angle traced by minute hand in 10 min = 10 × 6° = 60°

∴ required angle = 125° - 60° = 65°

Answer: (d)

Angle traced by hour hand in 35 min after $8 = 35 × 1°/2 = 17.5°$

At 8 : 35, min hand is at 7, and angle between 8 and 7 = 30°

∴ required angle between two hand at 8 : 35 = 30° + 17.5° = 47.5°

Answer: (d)

Angle traced by the minute hand in 2 hrs 20 min,

i.e., 140 min = $(360/60 × 140)^o= 840°$

Answer: (c)

∴ Angle traced by hour hand per minute = $(1/2)^o$

&there4 Angle traced by hour hand in 12 h 55 min (i.e., at 0 : 55) = $55 × 1°/2 = 27 1°/2$

&there4 angle traced by minute hand per minute = 6°

Therefore traced by minute hand in 55 min = 55 × 6° = 330°

Therefore required angle = $360° - (330° - 27 1°/2) = 360° + 27 1°/2 - 330°$

$= 387 1°/2 - 330° = 57 1°/2$

Answer: (c)

∴ Angle traced by hour hand per minute = $(1/2)^o$

∴ Angle traced by hour hand in 9 h 30 min

= [(9 × 60) + 30] × $1°/2$ = 570 × $1°/2$ = 285°

∴ angle traced by minute hand per minute = 6°

∴ angle traced by minute hand in 30 min = 30 × 6° = 180°

∴ required angle = (285° - 180°) = 105°

Answer: (b)

Angle traced by hour hand in $13/3 hrs = (360/12 × 13/3)^o= 130°$

Angle traced by min. hand in $20 min = (360/60 × 20)^o = 120°$

∴ Required angle = (130° - 120°) = 10°