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.

Result

Q =  46.597 °

Solution:

R=107+2260=322130≐107.3667 ∘ r=356 m q=271 m  sin⁡(Q)sin⁡(R)=qr  s=sin⁡(Q) s=qr⋅ sin⁡(R∘→ rad)=qr⋅ sin⁡(R∘⋅ π180 )=271356⋅ sin⁡(107.36666666667∘⋅ 3.1415926180 )=0.72653  Q=180∘π⋅arcsin⁡(s)=180∘π⋅arcsin⁡(0.7265)≐46.5966≐46.597∘≐46∘35′48"R=107+\dfrac{ 22 }{ 60 }=\dfrac{ 3221 }{ 30 } \doteq 107.3667 \ ^\circ \ \\ r=356 \ \text{m} \ \\ q=271 \ \text{m} \ \\ \ \\ \dfrac{ \sin(Q) }{ \sin(R) }=\dfrac{ q }{ r } \ \\ \ \\ s=\sin(Q) \ \\ s=\dfrac{ q }{ r } \cdot \ \sin( R ^\circ \rightarrow\ \text{rad})=\dfrac{ q }{ r } \cdot \ \sin( R ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ )=\dfrac{ 271 }{ 356 } \cdot \ \sin( 107.36666666667 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ )=0.72653 \ \\ \ \\ Q=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(s)=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(0.7265) \doteq 46.5966 \doteq 46.597 ^\circ \doteq 46^\circ 35'48"

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