Triangle $ABC$ where $a+c=12$, $b+c=13$, $\angle A = 60^{\circ}$, what is $a^{2} + b^{2} + c^{2}$?

Question

Triangle $ABC$ where $a+c=12$, $b+c=13$, $\angle A = 60^{\circ}$, what is $a^{2} + b^{2} + c^{2}$?

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By the cosine rule $$ a^{2} = b^{2} + c^{2} - bc$$

then since $(b+c)^{2}=169=b^{2}+c^{2}+2bc$, the above becomes

$$ a^{2} = 169 - 3bc$$

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We know $b = a+1$. Also

$$ (a+c)(b+c) = 156 $$ $$ ab + ac + bc + c^{2} = 156$$

Also, $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc)$, so $$ ab +ac + bc = \frac{(25-c)^{2}- (a^{2} + b^{2}+ c^{2})}{2}$$ $$ = \frac{(25-c)^{2}- (169 + b^{2}+ c^{2} - 3bc)}{2} $$ $$ = \frac{(25-c)^{2}- (169 + (13-c)^{2}+ c^{2} - 3(13-c)c)}{2} $$

How to solve this?

*Without Heron's formula?

Answer 1

Hint

$$b=a+1,c=12-a$$

$$2bc\cos A=b^2+c^2-a^2$$

Replace the values of $b,c,A=60^\circ$ to find $a$