F.4 MATHS TRIGONMETRY

2007-04-15 6:24 pm

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°.

(b)Given that θ=40, α=54 and β=46, find
(i) h/c,
(ii)the angle of elevation of B from Q,
(III) the angle of elevation and the bearing of B from M, where M is the mid-point of PQ.

回答 (1)

myisland8132

2007-04-16 2:14 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(100)
=61.29
h/c
=61.29/100
=0.6129
(ii)
QR/sinα=c/sin(α+β)
QR=csinα/sin(α+β)
QR=100sin54/sin100=82.1497
tan(the angle of elevation of B from Q)=BR/QR
tan(the angle of elevation of B from Q)=61.29/82.1497=0.746
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=73.0437
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=1.0298
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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