Observatory and aircraft

The aircraft flying towards the observatory was aimed at a distance of 5300 m at an elevation angle of 28º and after 9 seconds at a distance of 2400 m at an elevation angle of 50º. Calculate the distance the plane has flown in this time interval, its speed, and the change in altitude.

Correct answer:

x =  3203.5048 km
v =  355.945 m/s
h =  649.6926 m

Step-by-step explanation:

$$\begin{gathered} a=5300\text{m} \\ \alpha =28{}^{\circ } \\ t=9\text{s} \\ b=2400\text{m} \\ \beta =50{}^{\circ } \\ \sin \alpha =h_1:a \\ \cos \alpha =x_1:a \\ {h}_1=a\cdot \sin \alpha =a\cdot \sin 28°=5300\cdot \sin 28°=5300\cdot 0.469472=2488.19928\text{m} \\ {x}_1=a\cdot \cos \alpha =a\cdot \cos 28°=5300\cdot \cos 28°=5300\cdot 0.882948=4679.62224\text{m} \\ \sin \beta =h_2:b \\ \cos \beta =x_2:b \\ {h}_2=b\cdot \sin \beta =b\cdot \sin 50°=2400\cdot \sin 50°=2400\cdot 0.766044=1838.50666\text{m} \\ {x}_2=b\cdot \cos \beta =b\cdot \cos 50°=2400\cdot \cos 50°=2400\cdot 0.642788=1542.69026\text{m} \\ h=h_1-h_2=2488.1993-1838.5067\doteq 649.6926\text{m} \\ x=x_1-x_2=4679.6222-1542.6903\doteq 3203.5048\text{m} \\ x=\sqrt{h^2+x^2}=\sqrt{649.6926^2+3203.5048^2}\text{m}=3203.5048\text{km} \end{gathered}$$

$$\begin{gathered} v=\frac{x}{t}=\frac{3203.5048}{9}\doteq 355.945\text{m/s} \\ {v}_2=v\to \text{km/h}=v\cdot 3.6\text{km/h}=355.945\cdot 3.6\text{km/h}=1281.402\text{km/h} \end{gathered}$$

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