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

  • $\begingroup$ 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

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}


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 $(a,b,c)=(9,6,5)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.