In a triangle ABC, \(\angle\)ABC = 108° and \(\angle\)ACB = \(\frac{π}{10}\). The circular measure of \(\angle\)BAC is?
Answer & explanation
Correct answer: option 2
\(\angle\)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