Question

If the tangents of the angles of a triangle are in A.P prove that the squares of the sides are in the ratio${x^2}\left( {{x^2} + 9} \right):{\left( {3 + {x^2}} \right)^2}:9\left( {1 + {x^2}} \right)$, where x is tangents of the least or greatest angle.

Answer 1

Hint: Using$\operatorname{Tan} A\operatorname{Tan} B\operatorname{Tan} C = \tan A + \tan B + \tan C$, we will form the equations in x. Using these values we apply alternate sine rule to determine the value of squares of the sides. Alternate sine rule is$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}$, from this we can say$$\dfrac{{{a^2}}}{{{{\sin }^2}A}} = \dfrac{{{b^2}}}{{{{\sin }^2}B}} = \dfrac{{{c^2}}}{{{{\sin }^2}C}}$$.
Therefore$${a^2}:{b^2}:{c^2} = {\sin ^2}A:{\sin ^2}B:{\sin ^2}C$$

Complete step by step answer:
Given the tangents of the angles of a triangle are in Arithmetic Progression.
We need to prove that the squares of the sides are in the ratio${x^2}\left( {{x^2} + 9} \right):{\left( {3 + {x^2}} \right)^2}:9\left( {1 + {x^2}} \right)$
Arithmetic Progression: It is a sequence of numbers such that the difference of any two successive members is a constant.
Since$\tan A$,$\tan B$,$\tan C$are in A.P., we have
$\tan A + \tan C = 2\tan B$… (1)
Or $x + \tan C = 2\tan B$… (2)
Where x = tan A. (From the given: x is tangent of the least or greatest angle)
[Observer that if tan A, tan B, tan C are in A.P. then tan A is either the greatest of the least amongst tan A, tan B and tan C].
Now in a triangle ABC, we always have
$\operatorname{Tan} A\operatorname{Tan} B\operatorname{Tan} C = \tan A + \tan B + \tan C$… (3)
$\therefore $From (1), (2) and (3), we obtain
${\text{x tanB}}\left( {2\tan B - x} \right) = 2\tan B + \tan B$
${\text{x }}\left( {2\tan B - x} \right) = 3$ [Hence tan B$ \ne $0]
$\therefore $Tan B = $\dfrac{{\left( {3 + {x^2}} \right)}}{{2x}}$
And Tan C = 2 Tan B – x
$ \Rightarrow 2 \times \dfrac{{\left( {3 + {x^2}} \right)}}{{2x}} - x$
$ \Rightarrow \dfrac{{\left( {3 + {x^2}} \right)}}{x} - x$
$ \Rightarrow \dfrac{3}{x} + x - x$$ = \dfrac{3}{x}$
Now in a triangle $\vartriangle $ABC, we have
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}}$ (Alternate Sine rule)
$$ \Rightarrow \dfrac{{{a^2}}}{{{{\sin }^2}A}} = \dfrac{{{b^2}}}{{{{\sin }^2}B}} = \dfrac{{{c^2}}}{{{{\sin }^2}C}}$$
Hence $${a^2}:{b^2}:{c^2} = {\sin ^2}A:{\sin ^2}B:{\sin ^2}C$$
But tan A = x $$ \Rightarrow \sin A = \dfrac{x}{{\sqrt {1 + {x^2}} }}$$ (Using the trigonometric ratios and Pythagoras theorem)
$ \Rightarrow {\sin ^2}A = \dfrac{{{x^2}}}{{1 + {x^2}}}$
And tan B = $\dfrac{{\left( {3 + {x^2}} \right)}}{{2x}}$$$ \Rightarrow \sin B = \dfrac{{3 + {x^2}}}{{\sqrt {{{\left( {3 + {x^2}} \right)}^2} + 4{x^2}} }}$$
$$ \Rightarrow {\sin ^2}B = \dfrac{{{{\left( {3 + {x^2}} \right)}^2}}}{{{{\left( {\sqrt {{{\left( {3 + {x^2}} \right)}^2} + 4{x^2}} } \right)}^2}}}$$
$$ \Rightarrow {\sin ^2}B = \dfrac{{{{\left( {3 + {x^2}} \right)}^2}}}{{{{\left( {3 + {x^2}} \right)}^2} + 4{x^2}}}$$
$$ \Rightarrow {\sin ^2}B = \dfrac{{{{\left( {3 + {x^2}} \right)}^2}}}{{\left( {9 + 6{x^2} + {x^4}} \right) + 4{x^2}}}$$
$$ \Rightarrow {\sin ^2}B = \dfrac{{{{\left( {3 + {x^2}} \right)}^2}}}{{\left( {9 + 10{x^2} + {x^4}} \right)}}$$
$$ \Rightarrow {\sin ^2}B = \dfrac{{{{\left( {3 + {x^2}} \right)}^2}}}{{\left( {1 + {x^2}} \right)\left( {9 + {x^2}} \right)}}$$
And tan C = $\dfrac{3}{x}$$$ \Rightarrow \sin C = \dfrac{3}{{\sqrt {\left( {9 + {x^2}} \right)} }}$$
$$ \Rightarrow {\sin ^2}C = \dfrac{9}{{9 + {x^2}}}$$
$$\therefore {a^2}:{b^2}:{c^2} = {\sin ^2}A:{\sin ^2}B:{\sin ^2}C$$
$\therefore {a^2}:{b^2}:{c^2}::\dfrac{{{x^2}}}{{1 + {x^2}}}:\dfrac{{{{\left( {3 + {x^2}} \right)}^2}}}{{\left( {1 + {x^2}} \right)\left( {9 + {x^2}} \right)}}:\dfrac{9}{{9 + {x^2}}}::{x^2}\left( {9 + {x^2}} \right):{(3 + {x^2})^2}:9(1 + {x^2}).$

Note: We have an ambiguous case here that is, when two sides say ‘a’ and ‘b’ are given and one angle opposite to these sides say A is given, then by sine rule$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}}$. Therefore$\sin B = \dfrac{b}{a}\sin A = k$, say. There will be two values of angle B say${B_1}$and${B_2}$ which will satisfy the equation$\sin B = k$. Evidently these two values are supplementary i.e. if${B_1} = \alpha $then${B_2} = \pi - \alpha $i.e. ${B_1} + {B_2} = \pi $.