The triangle $ABC$ has a perimeter of 42 cm. Compute $x$ and $y$.

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So I used the Law of Cosines and found that $\alpha = {66,715}^o$ (in degrees). I don't know if I did the right thing because I'm stuck.

  • $\begingroup$ You must know that $x+y = 42-16-14=12$? Does that help? I think you'd be better off using radians, rather than degrees, too. $\endgroup$
    – amWhy
    Oct 3, 2016 at 13:30

2 Answers 2


You can use Angle Bisector Theorem. It states that for a triangle $ABC$ with the angle bisector from $A$ intersecting $BC$ at point $D$:

$$ \frac{AB}{AC} = \frac{BD}{DC} $$


$$ \frac{16}{14} = \frac{x}{y} $$

We also know that $x+y = 42 - 16 - 14 = 12$

From here we can derive: $$ x = 12 * \frac{16}{16+14} = \frac{192}{30} = 6\frac{2}{5} \\ y = 12 * \frac{14}{16+14} = \frac{168}{30} = 5\frac{3}{5} $$


It is possible to provide a solution starting from your computation which gives $2 \alpha = 66.715^{0}$ (beware: not $\alpha=66.715^{0}$), we can apply twice again the law of cosines in triangles BDA and DCA (denoting $L:=AD$).

$$\cases{x^2=16^2+L^2-2 \times L \times \cos(\alpha)\\y^2=14^2+L^2-2 \times L \times \cos(\alpha)}$$

If the second equation is substracted from the first, one gets an equation $x^2-y^2=k$ where $k$ is a known value. Otherwise said $(x+y)(x-y)=k$.

Besides, you know that $BC$ can be written in two ways: $x+y=12$.

With these two equations, it is easy to obtain the values of $x$ and $y$.


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