A) \[\frac{2y}{\sqrt{3x}}\]
B) \[\frac{x}{2y}\]
C) \[\frac{\sqrt{3x}}{2y}\]
D) \[\frac{2x}{\sqrt{3}y}\]
Correct Answer: C
Solution :
(c): From \[\Delta AQD,\] \[\sin {{60}^{{}^\circ }}=\frac{AQ}{AD}\] \[\Rightarrow \]\[\frac{\sqrt{3}}{2}=\frac{y}{AD}\] \[\Rightarrow \]\[AD=\frac{2y}{\sqrt{3}}\] From \[\Delta APD,\] \[sin{{75}^{{}^\circ }}=\frac{AP}{AD}=\frac{x}{\frac{2y}{\sqrt{3}}}=\frac{\sqrt{3}x}{2y}\]You need to login to perform this action.
You will be redirected in
3 sec