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:

$$\begin{gathered} AB=2000\text{m} \\ QAB=52°40^{\prime }=52°+\frac{40}{60}°=52.6667°\doteq 52.6667 \\ PBA=42°1^{\prime }=42°+\frac{1}{60}°=42.0167°\doteq 42.0167 \\ PAB=86°40^{\prime }=86°+\frac{40}{60}°=86.6667°\doteq 86.6667 \\ QBA=81°15^{\prime }=81°+\frac{15}{60}°=81.25°=81.25 \\ \angle APB=180-\angle PAB-\angle PBA=180-\frac{260}{3}-\frac{2521}{60}=\frac{3079}{60}\doteq 51.3167{}^{\circ } \\ \angle AQB=180-\angle QBA-\angle QAB=180-\frac{325}{4}-\frac{158}{3}=\frac{553}{12}\doteq 46.0833{}^{\circ } \\ PA:AB=\sin PBA:\sin APB \\ PA=AB\cdot \frac{\sin PBA}{\sin APB}=AB\cdot \frac{\sin 42.016666666667°}{\sin 51.316666666666°}=2000\cdot \frac{\sin 42.016666666667°}{\sin 51.316666666666°}=2000\cdot \frac{0.669347}{0.780612}=1714.92761\text{m} \\ QB=AB\cdot \frac{\sin QAB}{\sin AQB}=AB\cdot \frac{\sin 52.666666666667°}{\sin 46.083333333333°}=2000\cdot \frac{\sin 52.666666666667°}{\sin 46.083333333333°}=2000\cdot \frac{0.795121}{0.720349}=2207.59765\text{m} \\ PB=AB\cdot \frac{\sin PAB}{\sin APB}=AB\cdot \frac{\sin 86.666666666667°}{\sin 51.316666666666°}=2000\cdot \frac{\sin 86.666666666667°}{\sin 51.316666666666°}=2000\cdot \frac{0.998308}{0.780612}=2557.75683\text{m} \\ QA=AB\cdot \frac{\sin QBA}{\sin AQB}=AB\cdot \frac{\sin 81.25°}{\sin 46.083333333333°}=2000\cdot \frac{\sin 81.25°}{\sin 46.083333333333°}=2000\cdot \frac{0.988362}{0.720349}=2744.11706\text{m} \\ \angle PAQ=\angle PAB-\angle QAB=\frac{260}{3}-\frac{158}{3}=\frac{260-158}{3}=\frac{102}{3}=34{}^{\circ } \\ \angle PBQ=\angle QBA-\angle PBA=\frac{325}{4}-\frac{2521}{60}=\frac{4875}{60}-\frac{2521}{60}=\frac{4875-2521}{60}=\frac{2354}{60}=\frac{1177}{30}\doteq 39.2333{}^{\circ } \\ {x}^2=P{A}^2+Q{A}^2-2\cdot PA\cdot QA\cdot \cos PAQ \\ x=\sqrt{P{A}^2+Q{A}^2-2\cdot PA\cdot QA\cdot \cos PAQ}=\sqrt{P{A}^2+Q{A}^2-2\cdot PA\cdot QA\cdot \cos 34°}=\sqrt{1714.9276^2+2744.1171^2-2\cdot 1714.9276\cdot 2744.1171\cdot \cos 34°}=\sqrt{1714.9276^2+2744.1171^2-2\cdot 1714.9276\cdot 2744.1171\cdot 0.829038}=1633.498=1633.4982\text{m} \end{gathered}$$

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