In a triangle ABC, \(\angle\)ABC = 108° and \(\angle\)ACB = \(\frac{π}{10}\). The circular measure of \(\angle\)BAC is?

\(\frac{3π}{10}\) radian

\(\angle\)ABC = 108° = 108 × \(\frac{π}{180}\) = \(\frac{3π}{5}\)

In \(\Delta \) ABC ; \(\angle\)A + \(\angle\)B + \(\angle\)C = π

⇒ \(\angle\)BAC + \(\frac{3π}{5}\) + \(\frac{π}{10}\) = π

⇒ \(\angle\)BAC = π - \(\frac{3π}{5}\) - \(\frac{π}{10}\)

            = \(\frac{10-6-1}{10}\)π = \(\frac{3π}{10}\) radians