Using the Law of Sines to solve the triangle if \( \angle A=111^{\circ}, a=34, b=14 \). If no answer exists, enter DNE for all answers. \( \angle B \) is \( \square \) degrees \( \angle C \) is \( \square \) degrees Assume \( \angle A \) is opposite side \( a, \angle B \) is opposite side \( b \), and \( \angle C \) is opposite side \( c \).

Without using a calculator, compute the sine and cosine of \( 120^{\circ} \) by using the reference angle. What is the reference angle? \( \square \) degrees. In what quadrant is this angle? \( \square \) (answer \( 1,2,3 \), or 4 )

Without using a calculator, compute the sine and cosine of \( 120^{\circ} \) by using the reference angle. What is the reference angle? In what quadrant is this angle? \( \sin \left(120^{\circ}\right)=\square \) degrees. \( \cos \left(120^{\circ}\right)=\square \) (Type sqrt(2) for \( \sqrt{2} \) and sqrt(3) for \( \sqrt{3} \).)

\begin{tabular}{l} Resoluer el siguiente triangub. \\ \hline \( 5^{\circ} \) \\ \( \sin \alpha \)\end{tabular}\( \frac{B}{\operatorname{sen} \beta} \)

\begin{tabular}{l} (b) Using the formula \( \operatorname{Sin} 2 A=2 \operatorname{Sin} A \operatorname{Cos} A \) \\ solve the equation \( \sin 2 x-\cos x=0 \) \\ in the interval \( 0^{\circ} \leq x \leq 360^{\circ} \) \\ \hline\end{tabular}

\begin{tabular}{l} (b) Using the formula \( \operatorname{Sin} 2 A=2 \operatorname{Sin} A \operatorname{Cos} A \) \\ solve the equation \( \sin 2 x-\cos x=0 \) \\ in the interval \( 0^{\circ} \leq x \leq 360^{\circ} \) \\ \hline\end{tabular}