A) \[\sin \theta =\frac{BQd}{p}\]
B) \[\sin \theta =\frac{p}{BQd}\]
C) \[\sin \theta =\frac{Bp}{Qd}\]
D) \[\sin \theta =\frac{pd}{BQ}\]
Correct Answer: A
Solution :
[a] : A to D is part of circle with centre C and radius CD = r. \[mv=p=BQr\]or\[r=\frac{p}{BQ}\] \[\sin \theta =\frac{ED}{CD}=\frac{d}{r}=\frac{BQd}{p}\]You need to login to perform this action.
You will be redirected in
3 sec