# 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:

$CD=300 \ \text{m} \ \\ \angle BDC=43 \ ^\circ \ \\ \angle DCB=90 \ ^\circ \ \\ \ \\ δ=\angle BDC ^\circ \rightarrow\ \text{rad}=\angle BDC ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ =43 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ =0.75049 \ \\ \ \\ \tan δ=BC / CD \ \\ \tan δ=x / CD \ \\ \ \\ x=CD \cdot \ \tan ( δ )=300 \cdot \ \tan ( 0.7505 )=279.755 \ \text{m}$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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