Given:
△ABC,
AD⊥BC,
AC=26,
CD=10,
BC=42Now, In
△ACD,
AC2=CD2+AD2262=102+AD2AD2=576AD=24Now, In
△ABD,
BD=BC−CDBD=42−10BD=32AB2=BD2+AD2AB2=322+242AB2=1600AB=40Now,
6cosx−5cosy+8tany= 6secx−5secy+8tany
= 6(HB)−5(HB)+8(PB)
= 6(ACAD)−5(ABBD)+8(ADBD)
= 6(2624)−5(4032)+8(2432)
= 132−254+6
= 26−25+244
= 254
= 614
∴m=6