Clouds

From two points, A and B, on the horizontal plane, a forehead cloud was observed above the two points under elevation angles 73°20' and 64°40'. Points A and B are separated by 2830 m. How high is the cloud?

Final Answer:

x =  16261.6 m

Step-by-step explanation:

$$\begin{gathered} \alpha =73°20^{\prime }=73°+\frac{20}{60}°=73.3333°\doteq 73.3333 \\ \beta =64°40^{\prime }=64°+\frac{40}{60}°=64.6667°\doteq 64.6667 \\ {t}_1=\tan \alpha =\tan 73.333333333333°=3.340233=3.34023 \\ {t}_2=\tan \beta =\tan 64.666666666667°=2.112335=2.11233 \\ s=2830\text{m} \\ b-a=s \\ x=a\cdot t_1=b\cdot t_2 \\ a\cdot t_1=(s+a)\cdot t_2 \\ a=\frac{s\cdot t_2}{t_1-t_2}=\frac{2830\cdot 2.1123}{3.3402-2.1123}\doteq 4868.408\text{m} \\ b=s+a=2830+4868.408\doteq 7698.408\text{m} \\ x=a\cdot t_1=4868.408\cdot 3.3402\doteq 16261.6151\text{m} \\ \text{ Verifying Solution: } \\ A=\frac{180°}{\pi }\cdot \arctan (x/a)=\frac{180°}{\pi }\cdot \arctan (16261.6151/4868.408)=\frac{220}{3}\doteq 73.3333{}^{\circ } \\ B=\frac{180°}{\pi }\cdot \arctan (x/b)=\frac{180°}{\pi }\cdot \arctan (16261.6151/7698.408)=\frac{194}{3}\doteq 64.6667{}^{\circ } \end{gathered}$$

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