6
$\begingroup$

If in a triangle $\tan A:\tan B:\tan C = 1:2:3$ then, what are the ratio of the sides $a,b,c $?

$\endgroup$
7
$\begingroup$

As shown in this answer, if $A+B+C=\pi$, then $$ \tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)\tag{1} $$ Suppose $\tan(A)=x$, then $(1)$ becomes $$ 6x=(x+2x+3x)=(x\cdot2x\cdot3x)=6x^3\tag{2} $$ whose solutions are $x\in\{-1,0,1\}$. Therefore, $\tan(A)=1$, $\tan(B)=2$, and $\tan(C)=3$. Therefore, $$ \sin(A)=\frac1{\sqrt2},\sin(B)=\frac2{\sqrt5},\sin(C)=\frac3{\sqrt{10}}\tag{3} $$ and by the Law of Sines, those are the ratios of the sides opposite those angles.

$\endgroup$
2
$\begingroup$

Let $AD, BE, CF$ be altitudes of triangle $ABC$. Denote $BC=a, CA=b, AB=c$.

Observe that $$\frac 23=\frac{\tan B}{\tan C} = \frac{\frac{AD}{BD}}{\frac{AD}{DC}} = \frac{DC}{BD}$$ so $BD=\frac 35 a$ and $DC=\frac 25 a$.

On the other hand Pythagorean theorem says that $$c^2-b^2=AB^2-AC^2=(BD^2+AD^2)-(CD^2+AD^2)=BD^2-CD^2=\left(\frac 35 a\right)^2 - \left(\frac 25 a\right)^2 = \frac 15 a^2$$

Analogously we find $CE=\frac 14 b$, $AE=\frac 34 b$ and $$c^2-a^2=\frac 12 b^2$$

The above results imply that $b^2=\frac 85 a^2$ and $c^2=\frac 95 a^2$. Thus $a^2:b^2:c^2=5:8:9$ which leads to the answer: $$a:b:c=\sqrt 5 : 2\sqrt 2 : 3.$$

$\endgroup$
1
$\begingroup$

$$ sinA = \frac{a}{2R} $$ where R is the radius of the circumcircle. It'll be cancelled out later. $$ cosA = \frac{b^2+c^2-a^2}{2bc} $$ So, $$ tanA = \frac{abc}{R(b^2+c^2-a^2)} $$ And similarly, $$ tanB = \frac{abc}{R(a^2+c^2-b^2)},tanC = \frac{abc}{R(a^2+b^2-c^2} $$ Canceling out $\frac{abc}{R}$ from the ratio, we get, $$ \frac{1}{b^2+c^2-a^2}:\frac{1}{a^2+c^2-b^2}:\frac{1}{a^2+b^2-c^2} = 1:2:3 $$ Equating each of these terms to k, 2k and 3k respectively (so that the ratio 1:2:3 is maintained), we get the following equations $$ b^2+c^2-a^2 = \frac{1}{k} $$ $$ a^2+c^2-b^2 = \frac{1}{2k} $$ $$ a^2+b^2-c^2 = \frac{1}{3k} $$ On solving these, we get $a=\frac{\surd{5}}{2\surd{3}k}$, $ b=\frac{\surd{2}}{\surd{3}k} $ and $c=\frac{\surd{3}}{2k}$

Therefore $a:b:c = \surd{5}:2\surd{2}:3$

$\endgroup$

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.