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given: $\triangle ABC$ $P=20$ Note: P is perimeter $\cos \alpha = -\frac{1}{3}$ $\cos \beta = \frac{7}{9}$

Find the sides of the triangle I'm totally lost on this one. I have no idea from where to begin. The answer given in my textbook is: $a = 9 b = 6 c = 5$

I managed to solve the problem, I case somebody needs to know how: Find $\sin\alpha$ and $sin \beta$ Then $\gamma = 180 - (\alpha + \beta)$ => $sin \gamma = sin(\alpha + \beta)$ From sin law => a : b : c = $\sin \alpha : \sin \beta : \sin \gamma$ and a + b + c =20 from here it's just arithmetic operations

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  • $\begingroup$ For which vertices $\alpha$ and $\beta$ correspond? $\endgroup$ – sinbadh Jan 2 '16 at 13:00
  • $\begingroup$ I have give all the information which is presented in my text book. But my guess is a - $\alpha$ and b -$\beta$ $\endgroup$ – Planet_Earth Jan 2 '16 at 13:02
  • $\begingroup$ en.wikipedia.org/wiki/Law_of_sines $\endgroup$ – mathlove Jan 2 '16 at 13:10
  • $\begingroup$ $\sin \alpha=\frac{2\sqrt2}{3}; \sin \beta=\frac{4\sqrt2}{9}; \gamma=180-(\alpha+\beta);\cos \gamma=-(\cos\alpha\cos\beta-\sin\alpha\sin\beta);a+b+c=20;c=a\cos\beta+b\cos\alpha\iff c=\frac{7b}{9}-\frac a3; etc.......$ You get a system giving $a,b,c$ $\endgroup$ – Piquito Jan 2 '16 at 13:13
  • $\begingroup$ Yeah I managed to solve it already, thank you for your help. $\endgroup$ – Planet_Earth Jan 2 '16 at 13:13
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I'll assume that $a$ is the side opposite to $\alpha$ and similarly for $b$ and $c$, opposite to $\beta$ and $\gamma$.

The cosine law tells you that $$ a^2=b^2+c^2-2bc\cos\alpha $$ so $$ 2bc+2bc\cos\alpha=b^2+2bc+c^2-a^2=(b+c)^2-a^2=(a+b+c)(b+c-a) $$ which means $$ \frac{1}{15}bc=b+c-a $$ Similarly, $$ 2ac(1+\cos\beta)=(a+b+c)(a+c-b) $$ that is $$ \frac{8}{45}ac=a+c-b $$ Summing the two relations we get $$ \frac{c}{45}(8a+3b)=2c $$ that implies $8a+3b=90$ (because we assume $c\ne0$ in a triangle).

The sine law tells you that $$ \frac{a}{\sin\alpha}=\frac{b}{\sin\beta} $$ Since $$ \sin\alpha=\sqrt{1-\frac{1}{9}}=\frac{2}{3}\sqrt{2} $$ and $$ \sin\beta=\sqrt{1-\frac{49}{81}}=\frac{4}{9}\sqrt{2} $$ we have $$ \frac{3a}{2\sqrt{2}}=\frac{9b}{4\sqrt{2}} $$ that simplifies to $2a=3b$.

Now we have \begin{cases} 8a+3b=90\\[3px] 2a=3b\\[3px] a+b+c=20 \end{cases}

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Ok. I'll take $\alpha$ corresponds to vertice $A$ and $\beta$ to vertice $B$. Also, i´ll denonte by $a,b,c$ the oposite sides to vertices $A,B,C$ respectively.

By Cosine's Law, and using hypothesis about $P$, we have the system: $$\left\{ \begin{eqnarray} 20&=&a+b+c&\\ a^2&=&b^2+c^2+2bc\frac{1}{3}\\ b^2&=&c^2+a^2-2ac\frac{7}{9}\\ \end{eqnarray}\right.$$

Solving, we have three tripletes of solutions:

$(a,b,c)=(9,6,5)$ or $(a,b,c)=(10,10,0)$ or $(a,b,c)=(15,-10,15)$. But the last two tripletes can't be solution due to the fact that 0, in the second triplete, and -10, in the third triplete, can't be the measure of the sides of a triangle.

Then, the solution is $(a,b,c)=(9,6,5)$

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