The sides of a triangle are in A.P. and the greatest angle exceeds the least by $90^\circ$. Prove that the sides are proportional to $\sqrt{7} + 1$, $\sqrt{7}$ and $\sqrt{7} -1$.
Let $ABC$ be a triangle with sides $a, \; b, \; c$ and angles $A, \; B, \; C$.
Let $A$ be the least angle and $C$ the greatest angle.
Given: $\;$ $a, \; b, \; c$ are in A.P.
$\implies$ $2 b = a + c$ $\;\;\; \cdots \; (1a)$
and, $\;$ $C = A + 90^{\circ}$ $\;\;\; \cdots \; (1b)$
To prove that: $\;$ $a : b : c = \sqrt{7} + 1 : \sqrt{7} : \sqrt{7} - 1$
In $\triangle ABC$, we have by sine rule,
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
i.e. $\;$ $a = 2 R \sin A$ $\;\; \cdots \; (2a)$, $\;$ $b = 2 R \sin B$ $\;\; \cdots \; (2b)$, $\;$ $c = 2 R \sin C$ $\;\; \cdots \; (2c)$
In view of equations $(2a)$, $(2b)$ and $(2c)$, equation $(1a)$ becomes
$2 \times 2 R \sin B = 2 R \sin A + 2 R \sin C$
i.e. $\;$ $2 \sin B = \sin A + \sin C$ $\;\;\; \cdots \; (3a)$
$\left[\text{Note: } \sin \theta = 2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) \right.$
$\left. \sin \alpha + \sin \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha - \beta}{2}\right) \right]$
i.e. $\;$ $2 \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) = 2 \sin \left(\dfrac{A + C}{2}\right) \cos \left(\dfrac{A - C}{2}\right)$
i.e. $\;$ $2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) = \sin \left(\dfrac{A + C}{2}\right) \cos \left(\dfrac{A - C}{2}\right)$ $\;\;\; \cdots \; (3b)$
Now, in $\triangle ABC$, $\;$ $A + B + C = \pi$
$\implies$ $A + C = \pi - B$
$\implies$ $\dfrac{A + C}{2} = \dfrac{\pi}{2} - \dfrac{B}{2}$
$\therefore \;$ $\sin \left(\dfrac{A + C}{2}\right) = \sin \left[\dfrac{\pi}{2} - \left(\dfrac{B}{2}\right)\right] = \cos \left(\dfrac{B}{2}\right)$ $\;\;\; \cdots \; (4a)$
From equation $1b$, $\;$ $A - C = - 90^{\circ}$
$\therefore \;$ $\dfrac{A - C}{2} = - 45^{\circ}$
$\therefore \;$ $\cos \left(\dfrac{A - C}{2}\right) = \cos \left(- 45^{\circ}\right) = \dfrac{1}{\sqrt{2}}$ $\;\;\; \cdots \; (4b)$
$\therefore \;$ In view of equations $(4a)$ and $(4b)$, equation $(3b)$ becomes
$2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) = \dfrac{1}{\sqrt{2}} \times \cos \left(\dfrac{B}{2}\right)$
i.e. $\sin \left(\dfrac{B}{2}\right) = \dfrac{1}{2 \sqrt{2}}$ $\;\; \cdots \; (5a)$
From equation $(3a)$ we have,
$\sin A + \sin C = 2 \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right)$ $\;\;\; \cdots \; (6)$
We have from equation $(5a)$,
$\cos \left(\dfrac{B}{2}\right) = \sqrt{1 - \sin^2 \left(\dfrac{B}{2}\right)} = \sqrt{1 - \dfrac{1}{8}} = \dfrac{\sqrt{7}}{2 \sqrt{2}}$ $\;\;\; \cdots \; (5b)$
$\therefore \;$ In view of equations $(5a)$ and $(5b)$ equation $(6)$ becomes
$\sin A + \sin C = 4 \times \dfrac{1}{2 \sqrt{2}} \times \dfrac{\sqrt{7}}{2 \sqrt{2}}$
i.e. $\;$ $\sin A + \sin C = \dfrac{\sqrt{7}}{2}$ $\;\;\; \cdots \; (6a)$
$\left[\text{Note: } \sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha + \beta}{2}\right)\right]$
Now, $\;$ $\sin A - \sin C = 2 \sin \left(\dfrac{A - C}{2}\right) \cos \left(\dfrac{A + C}{2}\right)$ $\;\;\; \cdots \; (7)$
In $\triangle ABC$, $\;$ $\cos \left(\dfrac{A + C}{2}\right) = \cos \left[\dfrac{\pi}{2} - \left(\dfrac{B}{2}\right)\right] = \sin \left(\dfrac{B}{2}\right)$
i.e. $\;$ $\cos \left(\dfrac{A + C}{2}\right) = \dfrac{1}{2 \sqrt{2}}$ $\;\;$ [in view of equation $(5a)$] $\;\;\; \cdots \; (8a)$
From equation $(1b)$, $\;$ $\sin \left(\dfrac{A - C}{2}\right) = \sin \left(- 45^{\circ}\right) = \dfrac{-1}{\sqrt{2}}$ $\;\;\; \cdots \; (8b)$
$\therefore \;$ In view of equations $(8a)$ and $(8b)$ equation $(7)$ becomes
$\sin A - \sin C = 2 \times \left(\dfrac{-1}{\sqrt{2}}\right) \times \dfrac{1}{2 \sqrt{2}}$
i.e. $\;$ $\sin A- \sin C = \dfrac{-1}{2}$ $\;\;\; \cdots \; (7a)$
Adding equations $(6a)$ and $(7a)$ we have,
$2 \sin A = \dfrac{\sqrt{7} - 1}{2}$
$\implies$ $\sin A = \dfrac{\sqrt{7} - 1}{4}$ $\;\;\; \cdots \; (9a)$
Substituting the value of $\sin A$ in equation $(6a)$ we have,
$\sin C = \dfrac{\sqrt{7}}{2} - \dfrac{\sqrt{7} - 1}{4}$
$\implies$ $\sin C = \dfrac{\sqrt{7} + 1}{4}$ $\;\;\; \cdots \; (9b)$
We have from equations $(3a)$, $(9a)$ and $(9b)$,
$\sin B = \dfrac{\sin A + \sin C}{2}$
i.e. $\sin B = \dfrac{\dfrac{\sqrt{7} - 1}{4} + \dfrac{\sqrt{7} + 1}{4}}{2}$
i.e. $\;$ $\sin B = \dfrac{\sqrt{7}}{4}$ $\;\;\; \cdots \; (9c)$
Now, by sine rule, $\;$ $a : b : c = \sin A : \sin B : \sin C$
i.e. $\;$ $a : b : c = \dfrac{\sqrt{7} - 1}{4} : \dfrac{\sqrt{7}}{4} : \dfrac{\sqrt{7} + 1}{4}$ $\;\;$ [by equations $(9a)$, $(9b)$, $(9c)$]
i.e. $\;$ $a : b : c = \sqrt{7} - 1 : \sqrt{7} : \sqrt{7} + 1$
Hence proved.