Inaccessible 69794

Determine the distance between two inaccessible places P, Q, if the distance between two observation points A, B is 2000m and if you know the size of the angles QAB = 52°40''; PBA = 42°01''; PAB = 86°40'' and QBA = 81°15''. The considered locations A, B, P, and Q lie in one plane.

Correct answer:

x =  1633.4982 m

Step-by-step explanation:

AB=2000 m QAB=52°40=52°+6040°=52.6667°52.6667 PBA=42°1=42°+601°=42.0167°42.0167 PAB=86°40=86°+6040°=86.6667°86.6667 QBA=81°15=81°+6015°=81.25°=81.25  APB=180PABPBA=1803260602521=60307951.3167  AQB=180QBAQAB=18043253158=1255346.0833   PA:AB = sin PBA: sin APB  PA=AB sinAPBsinPBA=AB sin51.316666666666° sin42.016666666667° =2000 sin51.316666666666° sin42.016666666667° =2000 0.7806120.669347=1714.92761 m  QB=AB sinAQBsinQAB=AB sin46.083333333333° sin52.666666666667° =2000 sin46.083333333333° sin52.666666666667° =2000 0.7203490.795121=2207.59765 m  PB=AB sinAPBsinPAB=AB sin51.316666666666° sin86.666666666667° =2000 sin51.316666666666° sin86.666666666667° =2000 0.7806120.998308=2557.75683 m  QA=AB sinAQBsinQBA=AB sin46.083333333333° sin81.25° =2000 sin46.083333333333° sin81.25° =2000 0.7203490.988362=2744.11706 m  PAQ=PABQAB=32603158=3260158=3102=34  PBQ=QBAPBA=4325602521=604875602521=6048752521=602354=30117739.2333   x2 = PA2 + QA2  2 PA QA cos PAQ  x=PA2+QA22 PA QA cosPAQ=PA2+QA22 PA QA cos34° =1714.92762+2744.117122 1714.9276 2744.1171 cos34° =1714.92762+2744.117122 1714.9276 2744.1171 0.829038=1633.498=1633.4982 m

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