Observation tower

From the observation tower at a height of 105 m above sea level, the ship is aimed at a depth angle of 1° 49´. How far is the ship from the base of the tower?

Result

x =  198.695 km

Solution:

h=105 m A=1/60+49/3600=10936000.0303   tanA=h:x  x1=h/tanA=h/tan0.0302777777778 =105/tan0.0302777777778 =h/0.000528=198695.43706  x=x1km=x1/1000 km=198695.437063/1000 km=198.69544 km=198.695 kmh=105 \ \text{m} \ \\ A=1/60 + 49/3600=\dfrac{ 109 }{ 3600 } \doteq 0.0303 \ ^\circ \ \\ \ \\ \tan A=h:x \ \\ \ \\ x_{1}=h/\tan A ^\circ =h/\tan 0.0302777777778^\circ \ =105/\tan 0.0302777777778^\circ \ =h/0.000528=198695.43706 \ \\ \ \\ x=x_{1} \rightarrow km=x_{1} / 1000 \ km=198695.437063 / 1000 \ km=198.69544 \ km=198.695 \ \text{km}



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