C-10-Q054mediumsingle_mcqThe distance xxx satisfies—ax=105⋅tan60∘x = 105 \cdot \tan 60^\circx=105⋅tan60∘bx=105tan60∘x = \dfrac{105}{\tan 60^\circ}x=tan60∘105cx=105⋅sin60∘x = 105 \cdot \sin 60^\circx=105⋅sin60∘dx=105sin60∘x = \dfrac{105}{\sin 60^\circ}x=sin60∘105