Answer
Picture (A).
For the point A:
$$\sin\alpha=0.918$$
$$\cos\alpha=0.3965$$
$$\tan\alpha\approx 2.315$$
For the point B:
$$\sin\angle B =0.3965, \cos\angle B = - 0.918, $$
$$\tan\angle B = -\dfrac{0.3965}{0.918}\approx - 0.432$$
For the point C:
$$\sin\angle C =-0.918, \cos\angle C = - 0.3965, $$
$$\tan\angle C = \dfrac{0.918}{0.3965}\approx 2.315$$
For the point D:
$$\sin\angle D =-0.3965, \cos\angle D = 0.918, $$
$$\tan\angle D = -\dfrac{0.3965}{0.918}\approx - 0.432$$
Picture (B):
For the point A:
$$\sin\alpha=0.918$$
$$\cos\alpha=0.3965$$
$$\tan\alpha\approx 2.315$$
For the point B:
$$\sin\angle B =0.918, \cos\angle B = - 0.3965, $$
$$\tan\angle B = -\dfrac{0.3965}{0.918}\approx - 2.315$$
For the point C
$$\sin\angle C =-0.918, \cos\angle C = - 0.3965, $$
$$\tan\angle C = \dfrac{0.918}{0.3965}\approx 2.315$$
For the point D
$$\sin\angle D =-0.918, \cos\angle D = 0.3965, $$
$$\tan\angle D = -\dfrac{0.918}{0.3965}\approx -2.315$$
Work Step by Step
Let's start with pic. (A)
For the point A:
$$\sin\alpha=y=0.918$$
$$\cos\alpha=x=0.3965$$
$$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{y}{x}=\dfrac{0.918}{0.3965}\approx 2.315$$
For the point B the angle (formally call it $\angle B$) is obvious $\dfrac{\pi}{2} + \alpha$.
For the point C the angle ($\angle C$) is $\pi + \alpha.$
And for the point D ($\angle D$) is $\dfrac{3\pi}{2}+\alpha.$
We can use the equalities:
$$\sin(\pi/2+\alpha)=\cos\alpha, \cos(\pi/2+\alpha)=-\sin\alpha$$
$$\tan(\pi/2+\alpha)=\dfrac{\sin(\pi/2+\alpha)}{ \cos(\pi/2+\alpha)}=-\dfrac{\cos\alpha}{\sin\alpha}$$
So
$$\sin\angle B =0.3965, \cos\angle B = - 0.918, $$
$$\tan\angle B = -\dfrac{0.3965}{0.918}\approx - 0.432$$
For the point C:
$$\sin(\pi+\alpha)=-\sin\alpha, \cos(\pi+\alpha)=-\cos\alpha$$
$$\tan(\pi+\alpha)=\dfrac{\sin(\pi+\alpha)}{ \cos(\pi+\alpha)}=\dfrac{\cos\alpha}{\sin\alpha}$$
And
$$\sin\angle C =-0.918, \cos\angle C = - 0.3965, $$
$$\tan\angle C = \dfrac{0.918}{0.3965}\approx 2.315$$
For the point D
$$\sin(3\pi/2+\alpha)=-\cos\alpha, \cos(3\pi/2+\alpha)=\sin\alpha$$
$$\tan(3\pi/2+\alpha)=\dfrac{\sin(3\pi/2+\alpha)}{ \cos(3\pi/2+\alpha)}=-\dfrac{\cos\alpha}{\sin\alpha}$$
And
$$\sin\angle D =-0.3965, \cos\angle D = 0.918, $$
$$\tan\angle D = -\dfrac{0.3965}{0.918}\approx - 0.432$$
Now let's deal with picture (B). Let's act in a similar way.
For the point A:
$$\sin\alpha=y=0.918$$
$$\cos\alpha=x=0.3965$$
$$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{y}{x}=\dfrac{0.918}{0.3965}\approx 2.315$$
For the point B the angle (formally call it $\angle B$) is obvious $\pi - \alpha$.
For the point C the angle ($\angle C$) is $\pi + \alpha.$
And for the point D ($\angle D$) is $2\pi-\alpha.$
We can use the equalities:
$$\sin(\pi-\alpha)=\sin\alpha, \cos(\pi-\alpha)=-\cos\alpha$$
$$\tan(\pi-\alpha)=\dfrac{\sin(\pi-\alpha)}{ \cos(\pi-\alpha)}=-\dfrac{\sin\alpha}{\cos\alpha}$$
So
$$\sin\angle B =0.918, \cos\angle B = - 0.3965, $$
$$\tan\angle B = -\dfrac{0.3965}{0.918}\approx - 2.315$$
For the point C:
$$\sin(\pi+\alpha)=-\sin\alpha, \cos(\pi+\alpha)=-\cos\alpha$$
$$\tan(\pi+\alpha)=\dfrac{\sin(\pi+\alpha)}{ \cos(\pi+\alpha)}=\dfrac{\cos\alpha}{\sin\alpha}$$
And
$$\sin\angle C =-0.918, \cos\angle C = - 0.3965, $$
$$\tan\angle C = \dfrac{0.918}{0.3965}\approx 2.315$$
For the point D
$$\sin(2\pi-\alpha)=-\sin\alpha, \cos(2\pi-\alpha)=\cos\alpha$$
$$\tan(2\pi-\alpha)=\dfrac{\sin(2\pi-\alpha)}{ \cos(2\pi-\alpha)}=-\dfrac{\sin\alpha}{\cos\alpha}$$
And
$$\sin\angle D =-0.918, \cos\angle D = 0.3965, $$
$$\tan\angle D = -\dfrac{0.918}{0.3965}\approx -2.315$$
