1.A balloon B is observed simultaneously from two points P and Q on a horizontal Ground. P is at a distance of c meters due north of Q. R is the projection of B on the ground . The bearings f the balloon from P and Q are Sα°E and Nβ°E respectively. The angle of elevation of B from P is θ°.
(a)If the height of the balloon is h metres, show that
h=(c*tanθ°*sinβ°)/〔sin(α°+β°0〕
f.4 maths trigonometry
2007-04-15 6:20 pm
回答 (1)
2007-04-16 2:15 am
✔ 最佳答案
(a)∠PRQ=180-α-β
PQ/sin∠PRQ=PR/sinβ
PR
=PQsinβ/sin∠PRQ
=csinβ/sin(α+β)
BR/PR=tanθ
BR
=PRtanθ
=ctanθsinβ/sin(α+β)
h=ctanθsinβ/sin(α+β)
(b)
(i)
sub θ=40, α=54 and β=46
h
=ctanθsinβ/sin(α+β)
=100tan40sin46/sin(1 00)
=61.29
h/c
=61.29/100
=0.6129
(ii)
QR/sinα=c/sin(α+β)
QR=csinα/sin(α+β)
QR=100sin54/sin100=8 2.1497
tan(the angle of elevation of B from Q)=BR/QR
tan(the angle of elevation of B from Q)=61.29/82.1497=0.7 46
the angle of elevation of B from Q=36.7
(iii)
PR/sinβ=c/sin(α+β)
PR=csinβ/sin(α+β)
PR=100sin46/sin100=7 3.0437
MR^2=PM^2+PR^2-2(PM)(PR)cosα
MR=59.5145
BR=61.29
tan(the angle of elevation of B from Q)=BR/MR
tan(the angle of elevation of B from Q)=61.29/59.5145=1.0 298
the angle of elevation of B from M=45.8
PR/sin∠PRQ=MR/sinα
73.0437/sin∠PRQ=59.5145/sin54
sin∠PRQ=0.9929
∠PRQ=83.18
So the bearing of B from M is N83.18E
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