MHB How Do You Find AC/AB Given Tan Ratios in a Triangle?

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In triangle ABC, the tangent ratios are given as tan A:tan B:tan C = 1:2:3. The problem requires finding the ratio AC/AB based on these tangent values. Members greg1313 and kaliprasad successfully provided correct solutions to the problem. The discussion emphasizes the importance of understanding the relationships between angles and sides in triangles through trigonometric ratios. The thread serves as a platform for sharing problem-solving strategies in geometry.
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Here is this week's POTW:

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In a triangle $ABC$, $\tan A:\tan B: \tan C=1:2:3$. Find $\dfrac{AC}{AB}$.

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Congratulations to the following members for their correct solution::)

1. greg1313
2. kaliprasad

Solution from greg1313:
Using the triple tangent identity

$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$

and the given ratios we have

$$6\tan A=6\tan^3A\Rightarrow\tan A=1$$

Now we construct a triangle with base $\overline{AB}$ and an altitude from $C$ to $\overline{AB}$ at $P$.
WLOG, let $\overline{AP}=\overline{CP}=1$. Then, to form a tangent of $2$ at $B$, let $\overline{BP}=\frac12$.
We will now verify that $\tan C=3$.

Using the Pythagorean theorem with $\overline{AB}=\frac32$, $\overline{AC}=\sqrt2$ and $\overline{BC}=\frac{\sqrt5}{2}$,

$$\frac54-x^2=\frac94-(\sqrt2-x)^2\Rightarrow x=\frac{\sqrt2}{4}$$

where $x$ is the distance between the foot of an altitude from $B$ to $\overline{AC}$ and $C$.
Again with Pythagoras, the altitude from $B$ is $\frac{3\sqrt2}{4}$ and the tangent at $C$ is indeed $3$.

As $\overline{AC}=\sqrt2$ and $\overline{AB}=\frac32$, the desired ratio is $\frac{2\sqrt2}{3}$.
 
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