If the angles of a triangle be in A.P. and the lengths of the greatest and the least sides be $24 \; cm$ and $16 \; cm$ respectively, find the length of the third side and the other angles.
Let the angles of $\triangle ABC$ be $A$, $B$ and $C$.
Given: $\;$ $A, \; B, \; C$ are in $A.P.$
Then the least side is `$a$' and the greatest side is `$c$'.
Given: $\;$ Least side $= a = 16 \; cm$; $\;$ greatest side $= c = 24 \; cm$
$\because \;$ $A, \; B, \; C$ are in $A.P.$ $\implies$ $2 B = A + C$ $\;\;\; \cdots \; (1)$
Now, in $\triangle ABC$, $\;$ $A + B + C = 180^\circ$
$\implies$ $A + C = 180^\circ - B$ $\;\;\; \cdots \; (2)$
$\therefore \;$In view of equation $(2)$, equation $(1)$ becomes
$2 B = 180^\circ - B$
i.e. $3 B = 180^\circ$ $\implies$ $B = 60^\circ$
By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$
$\begin{aligned}
i.e. \; b^2 & = c^2 + a^2 - 2 c a \cos B \\\\
& = 24^2 + 16^2 - \left(2 \times 24 \times 16 \times \cos 60^\circ\right) \\\\
& = 576 + 256 - 384 = 448
\end{aligned}$
$\therefore \;$ $b = \sqrt{448} = 21.17 \; cm$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
$\therefore \;$ We have, $\;$ $\dfrac{16}{\sin A} = \dfrac{21.17}{\sin 60^\circ}$
i.e. $\;$ $\sin A = \dfrac{16 \times \sqrt{3}}{21.17 \times 2} = 0.6545$
$\implies$ $A = \sin^{-1} \left(0.6545\right) = 40.8817^\circ = 40^\circ 52' 54.12''$
and, $\;$ $\dfrac{21.17}{\sin 60^\circ} = \dfrac{24}{\sin C}$
i.e. $\;$ $\sin C = \dfrac{24 \times \sqrt{3}}{21.17 \times 2} = 0.9818$
$\implies$ $C = \sin^{-1} \left(0.9818\right) = 79.0520^\circ = 79^\circ 3' 7.2''$
$\therefore \;$ $b = 21.17 \; cm$, $\;$ $A = 40^\circ 52' 54.12''$, $\;$ $B = 60^\circ$, $\;$ $C = 79^\circ 3' 7.2''$