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Question Number 166917 by ajfour last updated on 02/Mar/22

Commented by mr W last updated on 02/Mar/22

Commented by ajfour last updated on 02/Mar/22
${(2b−c)cos θ−b}^2 +(2b−c)^2 sin^2 θ =(b+c)^2 ...(i) {(2b−c)cos θ−a}^2 +{b(√2)−(2b−c)sin θ}^2 =(a+c)^2 .....(ii) subtracting (i)−(ii) b^2 −a^2 −2(2b−c)(b−a)cos θ +2(√2)b(2b−c)sin θ−2b^2 =(b+c)^2 −(a+c)^2 ⇒ (2b−c)b(2(√2)sin θ−2cos θ) =2b^2 +bc ...(I) Also from..(i) (2b−c)^2 −2b(2b−c)cos θ=c^2 +2bc ⇒ cos θ=((4b^2 −6bc)/(2b(2b−c)))=((2b−3c)/(2b−c)) Also from ..(ii) (2b−c)^2 −2(√2)b(2b−c)sin θ −2a(2b−c)cos θ=c^2 +2ac−2b^2 ⇒ sin θ=(((2b−c)^2 +2b^2 −c^2 −2ac−2a(2b−3c))/(2(√2)b(2b−c))) =((((√2)b−(c/( (√2)))))/(2b−c)) substituting for sin θ & cos θ in (I) (((4b−2c))/(2b−c))−((2(2b−3c))/(2b−c))=((2b+c)/(2b−c)) ⇒ 3c=2b there seems to be some error..$
${(2 b - c) \cos θ - b}^{2} + {(2 b - c)}^{2} \sin^{2} θ$ $= {(b + c)}^{2} . . . (i)$ ${(2 b - c) \cos θ - a}^{2}$ $+ {b \sqrt{2} - (2 b - c) \sin θ}^{2} = {(a + c)}^{2}$ $. . . . . (i i)$ $s u b t r a c t i n g (i) - (i i)$ $b^{2} - a^{2} - 2 (2 b - c) (b - a) \cos θ$ $+ 2 \sqrt{2} b (2 b - c) \sin θ - 2 b^{2}$ $= {(b + c)}^{2} - {(a + c)}^{2}$ $\Rightarrow (2 b - c) b (2 \sqrt{2} \sin θ - 2 \cos θ)$ $= 2 b^{2} + b c . . . (I)$ $A l s o f r o m . . (i)$ ${(2 b - c)}^{2} - 2 b (2 b - c) \cos θ = c^{2} + 2 b c$ $\Rightarrow \cos θ = \frac{4 b^{2} - 6 b c}{2 b (2 b - c)} = \frac{2 b - 3 c}{2 b - c}$ $A l s o f r o m . . (i i)$ ${(2 b - c)}^{2} - 2 \sqrt{2} b (2 b - c) \sin θ$ $- 2 a (2 b - c) \cos θ = c^{2} + 2 a c - 2 b^{2}$ $\Rightarrow$ $\sin θ = \frac{{(2 b - c)}^{2} + 2 b^{2} - c^{2} - 2 a c - 2 a (2 b - 3 c)}{2 \sqrt{2} b (2 b - c)}$ $= \frac{(\sqrt{2} b - \frac{c}{\sqrt{2}})}{2 b - c}$ $s u b s t i t u t i n g f o r \sin θ & \cos θ i n (I)$ $\frac{(4 b - 2 c)}{2 b - c} - \frac{2 (2 b - 3 c)}{2 b - c} = \frac{2 b + c}{2 b - c}$ $\Rightarrow 3 c = 2 b$ $t h e r e s e e m s t o b e s o m e e r r o r . .$
Commented by mr W last updated on 02/Mar/22

$s o m e t h i n g s e e m s w r o n g i n d e e d s i r .$ $i g o t i n a n o t h e r w a y (s e e Q 77681) :$ ${(\frac{1}{a} + \frac{1}{2 a} + \frac{1}{c} - \frac{1}{4 a})}^{2} = 2 (\frac{1}{a^{2}} + \frac{1}{4 a^{2}} + \frac{1}{c^{2}} + \frac{1}{16 a^{2}})$ $17 {(\frac{c}{a})}^{2} - 40 (\frac{c}{a}) + 16 = 0$ $\Rightarrow \frac{c}{a} = \frac{4 (5 - 2 \sqrt{2})}{17} \approx 0.511$
Commented by Tawa11 last updated on 03/Mar/22

$Great sir$
	Answered by ajfour last updated on 02/Mar/22

	${(2 b - a)}^{2} - a^{2} = {(a + b)}^{2} - {(b - a)}^{2}$ $s a y \frac{b}{a} = x \Rightarrow$ ${(2 x - 1)}^{2} - 1 = {(1 + x)}^{2} - {(x - 1)}^{2}$ $4 x^{2} - 8 x = 0 \Rightarrow x = 2$
	Answered by cherokeesay last updated on 02/Mar/22

	${(2 b - a)}^{2} - a^{2} = x^{2} = {(a + b)}^{2} - {(b - a)}^{2} \Rightarrow$ $b = 2 a \Leftrightarrow \frac{b}{a} = 2$
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