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Question Number 25895 by Mr eaay last updated on 16/Dec/17

T h e h a n d s o f a c l o c k h a v e l e n g t h 3 c m

a n d 5 c m . C a l c u l a t e t h e d i s t a n c e b e t w e e n

t h e t i p s o f t h e h a n d a t 4 “ 0^{″} c l o c k .

Answered by mrW1 last updated on 16/Dec/17

a n g l e b e t w e e n t h e h a n d s :

a t h h o u r s a n d m m i n u t e s

θ = (\frac{h + \frac{m}{60}}{12} - \frac{m}{60}) \times 360 ° = (\frac{h}{12} - \frac{11 m}{720}) \times 360 °

a t 4 : 00

\Rightarrow θ = \frac{4}{12} \times 360 = 120 °

d i s t a n c e :

d = \sqrt{l_{H}^{2} + l_{M}^{2} - 2 l_{H} l_{M} \cos θ}

\Rightarrow d = \sqrt{3^{2} + 5^{2} - 2 \times 3 \times 5 \times (- 0.5)} = 7 c m

a t 4 : 40

θ = (\frac{4}{12} - \frac{11 \times 40}{720}) \times 360 ° = - 100 °

\Rightarrow d = \sqrt{3^{2} + 5^{2} - 2 \times 3 \times 5 \times \cos (- 100 °)} = 6.26 c m

Commented by Mr eaay last updated on 16/Dec/17

p l s e x p a n d m o r e \dots . H o w d o u g e t t h e a n g l e \dots

Commented by mrW1 last updated on 16/Dec/17

a t 4 : 00 t h e h o u r h a n d i s a t p o s i t i o n “ 4^{″}

a n d t h e m i n u t e h a n d i s a t p o s i t i o n “ 0^{″} .

12 h o u r s m e a n 360 °, s o 4 h o u r s m e a n

120 ° .

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