Answer:
E
Step-by-step explanation:
The problem says that triangle BDC lies in the plane k, which means that whatever angle is formed by another point beyond this plane with any of the three segments that form BDC (BD, DC, and BC) is the same as the angle formed by the line connecting the point and the plane.
Here, we're given that AD⊥DC, which means AD forms a 90° angle with DC. Then, since DC is already on the plane, we already know for sure that AD is definitely perpendicular to plane k.
Thus, the answer is E (none of these).
Quadrant IV
at quadrant IV, sinA is negative value while cosA is positive value
Hopefully this photo helps
