Question

in angle ABC , if 1/a+b +1/b+c=3/a+b+cthen show that C=60

Answer 1

Given :

In angle ABC,

$\\ \red \bigstar \tt \: \frac{1}{a + b} + \frac{1}{b + c} = \frac{3}{a + b + c}$

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To Prove :

• C =60 degrees

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Formulas :

By cosine Rule :

$\to\tt \: {c}^{2} = a + b - 2abcos \theta$

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Solution :

In a given ABC

$\\ \implies \sf \: \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c} \\ \\ \\ \underline { \tt {\bullet \: by \: cross \: multiplication : }} \\ \\ \\ \implies \sf \: \frac{a + b + 2c}{(a + c)(b + c)} = \frac{3}{a + b + c} \\ \\ \\ \implies \sf \: (a + b + 2c)(a + b + c) = 3(a + c)(b + c) \\ \\ \\ \implies \sf \: {a}^{2} + ab + ac + ab + {b}^{2} + bc + 2ac + 2bc + 2 {c}^{2} = 3\bigg(ab + ac + bc + {c}^{2} \bigg) \\ \\ \\ \implies \sf \: {a}^{2} + 2ab + 3ac + {b}^{2} + 3bc + 2 {c}^{2} = 3ab + 3ac + 3bc + 3 {c}^{2} \\ \\ \\ \implies \sf \: {a}^{2} + {b}^{2} = ab + {c}^{2} \\ \\ \\ \implies \sf \: {a}^{2} + {b}^{2} - {c}^{2} = ab \\ \\ \\ \implies \sf \: \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{ab} = 1 \\ \\ \\ \implies \sf \: \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} = \frac{1}{2} \\ \\ \\ \implies \sf \: cos \: c = cos 60 \degree \\ \\ \\ \implies \boxed{ \sf{c = 60 \degree}}$

•The value of c is 60 degree .

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Formulas :

Consider a triangle ABC with sides a, b, and c.

Sine Rule:

• For Finding sides :

$\\ \red \bigstar\sf{\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}}$

• For Finding Angles :

$\\ \red\bigstar\sf{\dfrac{sinA}{a}=\dfrac{sinB}{b}=\dfrac{sinC}{c}}$

Cosine Rule:

• For finding sides :

$\\ \pink\bigstar\sf{a^{2}=b^{2}+c^{2}-bccos(A)}$

• For finding Angles :

$\\ \pink\bigstar\sf{cos(A)=\dfrac{b^{2}+c^{2}-a^{2}}{2bc}}$