A drone

A flying drone aimed the area for an architect. He took off perpendicularly from point C to point D. He was at a height of 300 m above the plane of ABC. The drone from point D pointed at a BDC angle of 43°. Calculate the distance between points C and B in meters.

Correct result:

x =  279.7545 m

Solution:

$$\begin{gathered} CD=300\text{ m } \\ \mathrm{\angle }BDC=43^{\circ } \\ \mathrm{\angle }DCB=90^{\circ } \\ \delta =\mathrm{\angle }BD{C}^{\circ }\to \text{ rad}=\mathrm{\angle }BD{C}^{\circ }\cdot {\displaystyle \frac{\pi }{180}}=43^{\circ }\cdot {\displaystyle \frac{3.1415926}{180}}=0.75049 \\ \tan \delta =BC\mathrm{/}CD \\ \tan \delta =x\mathrm{/}CD \\ x=CD\cdot \tan (\delta )=300\cdot \tan (0.7505)=279.7545\text{ m} \end{gathered}$$

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