Encircled the given parts for easy reference
To solve for angle A, use SIN-TAAD rule for b
$\sin b = \tan \bar{A} ~ \tan a$
$\sin 62^\circ = \tan (90^\circ - A) ~ \tan 73^\circ$
$\tan (90^\circ - A) = \dfrac{\sin 62^\circ}{\tan 73^\circ}$
$90^\circ - A = 15.11^\circ$
$A = 74.89^\circ$ answer
To solve for side c, use SIN-COOP rule for $\bar{c}$
$\sin \bar{c} = \cos a ~ \cos b$
$\sin (90^\circ - c) = \cos 73^\circ ~ \cos 62^\circ$
$90^\circ - c = 7.89^\circ$
$c = 82.11^\circ$ answer
To solve for angle B, use SIN-TAAD rule for a
$\sin a = \tan b ~ \tan \bar{B}$
$\sin 73^\circ = \tan 62^\circ ~ \tan (90^\circ - B)$
$tan (90^\circ - B) = \dfrac{\sin 73^\circ}{\tan 62^\circ}$
$90^\circ - B = 26.95^\circ$
$B = 63.05^\circ$ answer
