Having had recent knee surgery, Tom is under doctor's orders not to hike up any incline that is $45^\circ$ or steeper.

(a) If $\theta$ is the angle formed by a hill and the ground such that $\tan(\theta)=$4.7, can Tom hike this hill? There may be more than one correct answer.

A. No. Since $\tan(\theta)=$4.7 and $\theta$ is an angle between 0 and $90^\circ$, then $\theta=\arctan$(4.7) is an angle greater than $45^\circ$.
B. No. Since $\tan(\theta)=$4.7 is positive, then $\theta$ must be an angle greater than $45^\circ$.
C. No. Since $\tan(\theta)=$4.7 and $\theta$ is an angle between 0 and $90^\circ$, then $\theta=\arctan$(4.7) is an angle less than $45^\circ$.
D. Yes. Since $\tan(\theta)=$4.7 and $\tan$(-1.78043616907932)$=\theta$, then $\theta=\arctan$(4.7)=-1.78043616907932 radians is an angle less than $\frac{\pi}{4}$ (or $45^\circ$).
E. None of the above

(b) What if $\sin(\theta)= 0.9$? Can Tom hike this hill? There may be more than one correct answer.

A. No. Since $\sin(\theta)=$0.9 is positive, then $\theta$ must be an angle greater than $45^\circ$.
B. Yes. Since $\sin(\theta)=$0.9 and $\sin$(-4.26136216499863)=0.9, then $\theta=\arcsin$(0.9)=-4.26136216499863 radians is an angle less than $\frac{\pi}{4}$ (or $45^\circ$).
C. Yes. Since $\sin(\theta)=$0.9 and $\theta$ is an angle between 0 and $90^\circ$, then $\theta=\arcsin$(0.9) is an angle less than $45^\circ$.
D. Yes. Since $\sin(\theta)=$0.9 is positive, then $\theta$ must be an angle between 0 and $45^\circ$.
E. No. Since $\sin(\theta)=$0.9 and $\theta$ is an angle between 0 and $90^\circ$, then $\theta=\arcsin$(0.9) is an angle greater than $45^\circ$.
F. None of the above

Hint: