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.You need to login to perform this action.
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