ab+c = bc+aab+c = bc+a\frac{a}{{b + c}}{\text{ = }}\frac{b}{{c + a}} = ca+b = k," role="presentation">ca+b = k,ca+b = k,\frac{c}{{a + b}}{\text{ = k,}} then find the value of k is" /> ab+c = bc+aab+c = bc+a\frac{a}{{b + c}}{\text{ = }}\frac{b}{{c + a}} = ca+b = k," role="presentation">ca+b = k,ca+b = k,\frac{c}{{a + b}}{\text{ = k,}} then find the value of k is" /> ab+c = bc+aab+c = bc+a\frac{a}{{b + c}}{\text{ = }}\frac{b}{{c + a}} = ca+b = k," role="presentation">ca+b = k,ca+b = k,\frac{c}{{a + b}}{\text{ = k,}} then find the value of k is" />

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If $\frac{a}{b + c} = \frac{b}{c + a}$ $\frac{a}{{b + c}}{\text{ = }}\frac{b}{{c + a}}$ = $\frac{c}{a + b} = k,$ $\frac{c}{{a + b}}{\text{ = k,}}$ then find the value of k is

1 $- \frac{1}{2}$ $- \frac{1}{2}$
21
3-1
4 $\frac{1}{2}$ $\frac{1}{2}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{a}{b + c} = k \\ \Rightarrow a = k (b + c) . . . . (i) \\ \frac{b}{c + a} = k \\ \Rightarrow b = k (c + a) . . . . (i i) \\ \frac{c}{a + b} = k \\ \Rightarrow c = k (a + b) . . . . (i i i) \\ Adding (i), (ii) and (iii) we get, \\ a + b + c \\ = k (2 a + 2 b + 2 c) \\ \Rightarrow k = \frac{a + b + c}{2 (a + b + c)} \\ = \frac{1}{2} \end{aligned}

$\eqalign{ & \frac{a}{{b + c}} = k \cr & \Rightarrow a = k\left( {b + c} \right)\,....(i) \cr & \frac{b}{{c + a}} = k \cr & \Rightarrow b = k\left( {c + a} \right)\,....(ii) \cr & \frac{c}{{a + b}} = k \cr & \Rightarrow c = k\left( {a + b} \right)\,....(iii) \cr & {\text{Adding (i), (ii) and (iii) we get,}} \cr & a + b + c \cr & = k\left( {2a + 2b + 2c} \right) \cr & \Rightarrow k = \frac{{a + b + c}}{{2(a + b + c)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr}$

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=

\frac{c}{a + b} = k,

$\frac{c}{{a + b}}{\text{ = k,}}$ then find the value of k is", "text": "If

\frac{a}{b + c} = \frac{b}{c + a}

$\frac{a}{{b + c}}{\text{ = }}\frac{b}{{c + a}}$ =

\frac{c}{a + b} = k,

$\frac{c}{{a + b}}{\text{ = k,}}$ then find the value of k is", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" }, "answerCount": "4", "acceptedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{a}{b + c} = k \\ \Rightarrow a = k (b + c) . . . . (i) \\ \frac{b}{c + a} = k \\ \Rightarrow b = k (c + a) . . . . (i i) \\ \frac{c}{a + b} = k \\ \Rightarrow c = k (a + b) . . . . (i i i) \\ Adding (i), (ii) and (iii) we get, \\ a + b + c \\ = k (2 a + 2 b + 2 c) \\ \Rightarrow k = \frac{a + b + c}{2 (a + b + c)} \\ = \frac{1}{2} \end{aligned}

$\eqalign{ & \frac{a}{{b + c}} = k \cr & \Rightarrow a = k\left( {b + c} \right)\,....(i) \cr & \frac{b}{{c + a}} = k \cr & \Rightarrow b = k\left( {c + a} \right)\,....(ii) \cr & \frac{c}{{a + b}} = k \cr & \Rightarrow c = k\left( {a + b} \right)\,....(iii) \cr & {\text{Adding (i), (ii) and (iii) we get,}} \cr & a + b + c \cr & = k\left( {2a + 2b + 2c} \right) \cr & \Rightarrow k = \frac{{a + b + c}}{{2(a + b + c)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} \frac{a}{b + c} = k \\ \Rightarrow a = k (b + c) . . . . (i) \\ \frac{b}{c + a} = k \\ \Rightarrow b = k (c + a) . . . . (i i) \\ \frac{c}{a + b} = k \\ \Rightarrow c = k (a + b) . . . . (i i i) \\ Adding (i), (ii) and (iii) we get, \\ a + b + c \\ = k (2 a + 2 b + 2 c) \\ \Rightarrow k = \frac{a + b + c}{2 (a + b + c)} \\ = \frac{1}{2} \end{aligned}

$\eqalign{ & \frac{a}{{b + c}} = k \cr & \Rightarrow a = k\left( {b + c} \right)\,....(i) \cr & \frac{b}{{c + a}} = k \cr & \Rightarrow b = k\left( {c + a} \right)\,....(ii) \cr & \frac{c}{{a + b}} = k \cr & \Rightarrow c = k\left( {a + b} \right)\,....(iii) \cr & {\text{Adding (i), (ii) and (iii) we get,}} \cr & a + b + c \cr & = k\left( {2a + 2b + 2c} \right) \cr & \Rightarrow k = \frac{{a + b + c}}{{2(a + b + c)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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