arithmetic sequence in a triangle

**metlx** · December 13th 2008, 12:36 PM

In what ratio are the sides of a triangle if one of the angles is 120° and the sides of the triangle make up a arethmetic sequence..

so we have 3 sides:
a = a
b = a +d
c = a +2d
and one of the angles is 120° = angle between a and a+d (the shortest sides)..

i tried a lot of things but i couldnt find a way to get the right result. =(

**Soroban** · December 13th 2008, 02:03 PM

Hello, metlx!

In what ratio are the sides of a triangle if one of the angles is 120°
and the sides of the triangle make up a arethmetic sequence?

so we have 3 sides: a = a, b = a +d , c = a +2d . . . . Good!

and one of the angles is 120° = angle between a and a+d (the shortest sides). . Right!

We need the Law of Cosines: . $c^2 \;=\;a^2 + b^2 - 2ab\cos C$

So we have: . $(a+2d)^2 \;=\;a^2 + (a+d)^2 - 2(a)(a+d)\cos120^o$

Since $\cos120^o = \text{-}\tfrac{1}{2}$ , we have: . $a^2 + 4ad + 4d^2 \;=\;a^2 + a^2 + 2ad + d^2 + a^2+ad$

. . which simplifies to: . $2a^2 - ad - 3d^2\:=\:0 \quad\Rightarrow\quad (a + d)(2a-3d) \:=\:0$

And we get: . $\begin{Bmatrix}a + d \:=\:0 & \Rightarrow & a \:=\:\text{-}d & \text{not possible} \\<br /> 2a-3d \:=\:0 & \Rightarrow & d \:=\:\frac{2}{3}a \end{Bmatrix}$

The shortest side is: . $a$

The next side is: . $b \:=\:a+d \:=\:a+\tfrac{2}{3}a \:=\:\tfrac{5}{3}a$

The longest side is: . $c \:=\:a + 2d \:=\:a + \tfrac{4}{3}a \:=\:\tfrac{7}{3}a$

The ratio of the sides is: . $a : \tfrac{5}{3}a : \tfrac{7}{3}a$

. . Multiply by $3\!:\;\;3a:5a:7a$

Therefore, the ratio of the sides is: . ${\color{blue}3:5:7}$

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