geometry or algebra

Geometry Level 2

$a,b,c$ denote sides opposite to angles $\alpha, \beta, \gamma$ . Given that $\frac { 1 }{ b+c } +\frac { 1 }{ a+c } =\frac { 3 }{ a+b+c }$ , then find angle $\gamma$ .

The answer is 60.

1 solution

Akash Deep
Aug 2, 2014

$\frac { 1 }{ b+c } +\frac { 1 }{ a+c } =\frac { 3 }{ a+b+c } \quad after\quad solving\quad we\quad get\\ (a+b+c)(a+b+c+c)\quad =\quad 3(b+c)(a+c)\\ on\quad simplification\quad we\quad get\quad \\ { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }\quad =\quad ab\\ multiplying\quad both\quad sides\quad by\quad 2\quad we\quad get\\ 2({ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 })\quad =\quad 2ab\\ 2\left\{ \frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab } \right\} =\quad 1\\ by\quad cosine\quad rule\quad ,\quad cos\quad c\quad =\quad \frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab } \\ 2\quad cos\quad c\quad =\quad 1\\ cos\quad c\quad =\quad \frac { 1 }{ 2 } \quad so\quad angle\quad c\quad =\quad 60\\$

Note that $a=b=c=1$ works, so the answer is $60$ .

Daniel Liu - 6 years, 10 months ago

actually a=b=c would do :D

Ko Io - 6 years, 10 months ago

Yes....but tell me that how would you find that a=b=c=1

Rahul Kumar - 6 years, 10 months ago

For that you need to look into @akash deep 's $solution$ . :P

If you still can't understand .....write back

Arya Samanta - 6 years, 10 months ago

Gamma is not specified.therefore alpha must be=to gamma= beta= 60

Godwin Tom George - 6 years, 10 months ago

geometry or algebra

The answer is 60.

1 solution

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