# 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 =  279.755 m

#### Solution:

$CD=300 \ \text{m} \ \\ \angle BDC=43 \ ^\circ \ \\ \angle DCB=90 \ ^\circ \ \\ \ \\ δ=\angle BDC ^\circ \rightarrow\ \text{rad}=\angle BDC ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ =43 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ =0.75049 \ \\ \ \\ \tan δ=BC / CD \ \\ \tan δ=x / CD \ \\ \ \\ x=CD \cdot \ \tan ( δ )=300 \cdot \ \tan ( 0.7505 )=279.755 \ \text{m}$

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