MATHS 28-31

28)CD=h/√3m
BD=h/√3+100m
angleA=45 degrees
therefore,h=h/√3+100
h=50(3+√3)m
29a)cosA+cosB+sinA+sinB/sinA+sinB(cosAcos(90-A)=sinAsin(90-A))
=1+1=2
30)horizontal distance =√64^2-25^2
=√3471
=58.915....m
b)25/58.91
=1/2.35661
n=2.35661
31)tanβ=5/7
β=35.53767779
the reduced bearing of C from A =S54.46W

28-31
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