The correct option is
C 9Draw a perpendicular from
D on
AB to meet at
AB at
E. Hence,
DEBC is a rectangle.
Thus,
DE=BC=12In
△DBC,
BD2=CD2+BC2132=122+CD2CD2=25CD=5Now,
BE=CD=5Thus,
AE=AB−BEAE=14−5AE=9Now, In
△AEDsinθ=DEAD=12ADAD=12cosecθcosΘ=AEAD=9ADAD=9secθ