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?

Correct result:

x =  198.6954 km

Solution:

$$\begin{gathered} h=105\text{ m } \\ A=1\mathrm{/}60+49\mathrm{/}3600={\displaystyle \frac{109}{3600}}\doteq 0.0303^{\circ } \\ \tan A=h:x \\ {x}_1=h\mathrm{/}\tan A^{\circ }=h\mathrm{/}\tan 0.0302777777778^{\circ }=105\mathrm{/}\tan 0.0302777777778^{\circ }=105\mathrm{/}0.000528=198695.43706\text{ m } \\ x=x_1\to \text{ km}=x_1\mathrm{/}1000\text{ km}=198695.4371\mathrm{/}1000\text{ km}=198.695\text{ km}=198.6954\text{ km} \end{gathered}$$

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