The lengths of two sides of a triangle are $1$ and $\sqrt{2}$ units respectively and the angle opposite the shorter side is $30^\circ$. Prove that there are two triangles satisfying these conditions, find their angles and show that their areas are in the ratio $\sqrt{3} + 1 : \sqrt{3} - 1$
Let the sides of the triangle be $\;$ $a = 1$ unit, $\;$ $b = \sqrt{2}$ units
Angle opposite shorter side is $\;$ $A = 30^\circ$
Now, $\;$ $b \sin A = \sqrt{2} \times \sin \left(30^\circ\right) = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}} < a$
$\implies$ There are two values of $B$, $\;$ $0^\circ < B < 90^\circ$ (first quadrant) and $90^\circ < B < 180^\circ$ (second quadrant)
By sine rule in $\triangle ABC$, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\implies$ $\sin B = \dfrac{b \sin A}{a} = \dfrac{\sqrt{2} \times \sin \left(30^\circ\right)}{1} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$
$\implies$ $B = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right)$
i.e. $\;$ $B = 45^\circ$ $\;$ or $\;$ $B = 180^\circ - 45^\circ = 135^\circ$
Now, in $\triangle ABC$, $A + B + C = 180^\circ$ $\implies$ $C = 180^\circ - \left(A + B\right)$
When $\;$ $B = 45^\circ$, $\;$ $C = 180^\circ - \left(30^\circ + 45^\circ\right) = 105^\circ$
When $\;$ $B = 135^\circ$, $\;$ $C = 180^\circ - \left(30^\circ + 135^\circ\right) = 15^\circ$
When $\;$ $a = 1$, $\;$ $b = \sqrt{2}$, $\;$ $A = 30^\circ$, $\;$ $B_1 = 45^\circ$, $\;$ $C_1 = 105^\circ$
Area of $\triangle A B_1 C_1$ $= \Delta_1 = \dfrac{1}{2} a \; b \sin C_1$
i.e. $\;$ $\Delta_1 = \dfrac{1}{2} \times 1 \times \sqrt{2} \times \sin \left(105^\circ\right)$
i.e. $\;$ $\Delta_1 = \dfrac{\sqrt{2}}{2} \times \dfrac{\left(\sqrt{3} + 1\right)}{2 \sqrt{2}}$
i.e. $\;$ $\Delta_1 = \dfrac{\sqrt{3} + 1}{4}$
When $\;$ $a = 1$, $\;$ $b = \sqrt{2}$, $\;$ $A = 30^\circ$, $\;$ $B_2 = 135^\circ$, $\;$ $C_2 = 15^\circ$
Area of $\triangle A B_2 C_2$ $= \Delta_2 = \dfrac{1}{2} a \; b \sin C_2$
i.e. $\;$ $\Delta_2 = \dfrac{1}{2} \times 1 \times \sqrt{2} \times \sin \left(15^\circ\right)$
i.e. $\;$ $\Delta_2 = \dfrac{\sqrt{2}}{2} \times \dfrac{\left(\sqrt{3} - 1\right)}{2 \sqrt{2}}$
i.e. $\;$ $\Delta_2 = \dfrac{\sqrt{3} - 1}{4}$
$\therefore \;$ $\Delta_1 : \Delta_2 = \dfrac{\sqrt{3} + 1}{4} : \dfrac{\sqrt{3} - 1}{4}$
i.e. $\;$ $\Delta_1 : \Delta_2 = \sqrt{3} + 1 : \sqrt{3} - 1$