SSA and geometry

The distance between the points P and Q was 356 m measured in the terrain. The PQ line can be seen from the viewer at a viewing angle of 107° 22 '. The observer's distance from P is 271 m. Determine the viewing angle of P and observer.

Correct result:

Q =  46.5966 °

Solution:

$$\begin{gathered} R=107+{\displaystyle \frac{22}{60}}={\displaystyle \frac{3221}{30}}\doteq 107.3667^{\circ } \\ r=356\text{ m } \\ q=271\text{ m } \\ {\displaystyle \frac{\sin (Q)}{\sin (R)}}={\displaystyle \frac{q}{r}} \\ s=\sin (Q) \\ s={\displaystyle \frac{q}{r}}\cdot \sin R^{\circ }={\displaystyle \frac{q}{r}}\cdot \sin 107.366666667^{\circ }={\displaystyle \frac{271}{356}}\cdot \sin 107.366666667^{\circ }={\displaystyle \frac{271}{356}}\cdot 0.954414=0.72653 \\ Q={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arcsin (s)={\displaystyle \frac{180^{\circ }}{\pi }}\cdot \arcsin (0.7265)=46.5966^{\circ }=46^{\circ }35^{\mathrm{\prime }}48\mathrm{"} \end{gathered}$$

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