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Question Number 30808 by 33 last updated on 26/Feb/18

	Answered by mrW2 last updated on 26/Feb/18

	Commented by mrW2 last updated on 26/Feb/18

	$x_{1} = 2 a \cos θ$ $x_{2} = x_{1} + 2 b \cos θ = (2 a + 2 b) \cos θ$ $\frac{x_{1}}{x_{2}} = \frac{2 a}{2 a + 2 b} = \frac{1}{1 + \frac{b}{a}} = \frac{1}{1 + \frac{2}{3}} = \frac{3}{5}$ $\frac{V_{1}}{V_{2}} = \frac{x_{1}}{x_{2}} = \frac{3}{5}$ $\Rightarrow V_{1} = \frac{3}{5} V_{2} = \frac{3}{5} V$
	Commented by 33 last updated on 26/Feb/18

	$y e s s i r t h i s w a s m y a n s w e r t o o .$
	Commented by 33 last updated on 26/Feb/18

	$b u t i c u d n t e x p l a i n w h y d i d w e$ $t a k e b o t h t h e a n g l e s t o b e s a m e .$ $i s t h a t a n a s s u m p t i o n$ $o r w e c a n p r o v e i t ?$
	Commented by rahul 19 last updated on 26/Feb/18

	$why x_{1} = 2 {a c o s}_{} θ ?$
	Commented by mrW2 last updated on 26/Feb/18

	$t w o r o d s a r e c o n n e c t e d a t p o i n t A_{1} .$ $x_{1} = A_{0} A_{1} = 2 \times a \cos θ$ $m e a n i n g o f a, b s e e m y d i a g r a m, n o t$ $t h e s a m e a s y o u r s .$
	Commented by mrW2 last updated on 26/Feb/18

	$I s a i d n o t 4 b u t 2 r o d s a r e c o n n e c t e d$ $a t p o i n t A_{1}, t h e r e f o r e t h e a n g l e i s t h e$ $s a m e, o t h e r w i s e p o i n t A_{1} h a s i n d e p e n d e n t$ $m o t i o n f r o m p o i n t A_{2} .$
	Commented by mrW2 last updated on 26/Feb/18

	$A_{0} i s f i x e d p o i n t, A_{1} i s n o t .$
	Commented by rahul 19 last updated on 26/Feb/18

	$ok, thank u sir .$
	Commented by 33 last updated on 26/Feb/18

	$s i r w h a t d o y o u s a y a b o u t$ $t h e a n g l e s ?$
	Commented by rahul 19 last updated on 26/Feb/18

	$ok, so u r taking distance from point A_{0} .$ $what if i take reference point as point A_{1} ?$
	Commented by 33 last updated on 26/Feb/18

	$o h y e s s i r . g o t i t t h a n k s$
	Answered by 33 last updated on 26/Feb/18

	Commented by rahul 19 last updated on 26/Feb/18

	$just a little mistake, although it {dosen}^{'} t$ $affect your calculations is that$ $l_{2} = 2 \times 3 a \cos θ and l_{1} = 2 \times 2 acos θ .$
	Answered by 33 last updated on 26/Feb/18

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