ICSE Class 9 Answered

The posted question is understood as given below

ΔABC is right triangle and right angle is at B . BD is perpendicular to opposite side BC.

What is DBC and what is sine of angle DBA ?

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DBC = 90° - ABD  ....................(1)

in right angle triangle ABD , DAB + ABD = 90° .................(2)

if we substitute ABD from eqn.(2) , then we rewrite eqn.(1) as

DBC =   DAB = CAB

sin( CAB) = BC/AC = 3 / = 3/5 = 0.6

CAB = sin^-1( 0.6 ) = 37°

Hence DBC = 37°

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DBA = ABC - DBC = 90° - 37° = 53°

sin(DBA) = sin53° = 0.8