The hands of a clock have length 3cm and 5cm.Calculate the distance between the tips of the hand at 4 “0” clock.


Question and Answers Forum

All Questions Topic List
Mensuration Questions
Previous in All Question Next in All Question
Previous in Mensuration Next in Mensuration
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

	$p l s e x p a n d m o r e . . . . H o w d o u g e t t h e a n g l e . . .$
	Commented by mrW1 last updated on 16/Dec/17

	$a t 4 : 00 t h e h o u r h a n d i s a t p o s i t i o n ‘ ‘ 4^{″}$ $a n d t h e m i n u t e h a n d i s a t p o s i t i o n ‘ ‘ 0^{″} .$ $12 h o u r s m e a n 360 °, s o 4 h o u r s m e a n$ $120 ° .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum