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Question Number 13738 by chux last updated on 22/May/17

P,Q,R,S are four locations on the same horizontal plane.Q is on a bearing of 041° from P and the distance is 40km. S is 28km from R on a bearing 074°, R is directly due north of P and the distance between Q and R is 38km. (a)the bearing of R from Q (b)the distance between Q and S (c)the distance between P and R

$$\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S}\:\mathrm{are}\:\mathrm{four}\:\mathrm{locations}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{horizontal}\:\mathrm{plane}.\mathrm{Q}\:\mathrm{is}\:\mathrm{on}\:\mathrm{a}\: \\ $$$$\mathrm{bearing}\:\mathrm{of}\:\mathrm{041}°\:\mathrm{from}\:\mathrm{P}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{is}\:\mathrm{40km}. \\ $$$$\mathrm{S}\:\mathrm{is}\:\mathrm{28km}\:\mathrm{from}\:\mathrm{R}\:\mathrm{on}\:\mathrm{a}\:\mathrm{bearing}\:\mathrm{074}°, \\ $$$$\mathrm{R}\:\mathrm{is}\:\mathrm{directly}\:\mathrm{due}\:\mathrm{north}\:\mathrm{of}\:\mathrm{P}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{R}\:\mathrm{is} \\ $$$$\mathrm{38km}. \\ $$$$\left(\mathrm{a}\right)\mathrm{the}\:\mathrm{bearing}\:\mathrm{of}\:\mathrm{R}\:\mathrm{from}\:\mathrm{Q} \\ $$$$\left(\mathrm{b}\right)\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{S} \\ $$$$\left(\mathrm{c}\right)\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{P}\:\mathrm{and}\:\mathrm{R} \\ $$

Commented by chux last updated on 23/May/17

please help

$$\mathrm{please}\:\mathrm{help} \\ $$

Answered by sandy_suhendra last updated on 23/May/17

Commented by sandy_suhendra last updated on 23/May/17

a) ((RQ)/(sin 41)) = ((PQ)/(sin θ)) ((38)/(0.656)) = ((40)/(sin θ)) sin θ = 0.691 ⇒ θ = 43.7° the bearing R from Q = 360−θ = 360−43.7 = 316.3° b) ∠QRS = 180−74−43.7=62.3° QS = (√(QR^2 +RS^2 −2QR.RS cos 62.3)) =(√(38^2 +28^2 −2×38×28×0.465)) =(√(1,238.48)) = 35.2 km c) ∠PQR = 180−41−43.7 = 95.3° PR=(√(QP^2 +QR^2 −2QP.QR.cos 95.3)) =(√(40^2 +38^2 −2×40×38×(−0.092))) =(√(3,323.68)) = 57.7 km

$$\left.\mathrm{a}\right)\:\frac{\mathrm{RQ}}{\mathrm{sin}\:\mathrm{41}}\:=\:\frac{\mathrm{PQ}}{\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\frac{\mathrm{38}}{\mathrm{0}.\mathrm{656}}\:=\:\frac{\mathrm{40}}{\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:\:\:\mathrm{sin}\:\theta\:=\:\mathrm{0}.\mathrm{691}\:\Rightarrow\:\theta\:=\:\mathrm{43}.\mathrm{7}° \\ $$$$\:\:\:\:\:\:\mathrm{the}\:\mathrm{bearing}\:\mathrm{R}\:\mathrm{from}\:\mathrm{Q}\:=\:\mathrm{360}−\theta\:=\:\mathrm{360}−\mathrm{43}.\mathrm{7}\:=\:\mathrm{316}.\mathrm{3}°\:\:\:\:\: \\ $$$$\left.\mathrm{b}\right)\:\angle\mathrm{QRS}\:=\:\mathrm{180}−\mathrm{74}−\mathrm{43}.\mathrm{7}=\mathrm{62}.\mathrm{3}° \\ $$$$\:\:\:\:\:\mathrm{QS}\:=\:\sqrt{\mathrm{QR}^{\mathrm{2}} +\mathrm{RS}^{\mathrm{2}} −\mathrm{2QR}.\mathrm{RS}\:\mathrm{cos}\:\mathrm{62}.\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{38}^{\mathrm{2}} +\mathrm{28}^{\mathrm{2}} −\mathrm{2}×\mathrm{38}×\mathrm{28}×\mathrm{0}.\mathrm{465}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{1},\mathrm{238}.\mathrm{48}}\:=\:\mathrm{35}.\mathrm{2}\:\mathrm{km} \\ $$$$\left.\mathrm{c}\right)\:\angle\mathrm{PQR}\:=\:\mathrm{180}−\mathrm{41}−\mathrm{43}.\mathrm{7}\:=\:\mathrm{95}.\mathrm{3}° \\ $$$$\:\:\:\:\:\mathrm{PR}=\sqrt{\mathrm{QP}^{\mathrm{2}} +\mathrm{QR}^{\mathrm{2}} −\mathrm{2QP}.\mathrm{QR}.\mathrm{cos}\:\mathrm{95}.\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{40}^{\mathrm{2}} +\mathrm{38}^{\mathrm{2}} −\mathrm{2}×\mathrm{40}×\mathrm{38}×\left(−\mathrm{0}.\mathrm{092}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{3},\mathrm{323}.\mathrm{68}}\:=\:\mathrm{57}.\mathrm{7}\:\mathrm{km} \\ $$

Commented by chux last updated on 23/May/17

i′m most grateful sir

$$\mathrm{i}'\mathrm{m}\:\mathrm{most}\:\mathrm{grateful}\:\mathrm{sir} \\ $$

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