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Question Number 92518 by I want to learn more last updated on 07/May/20

Answered by mr W last updated on 07/May/20

Commented by mr W last updated on 07/May/20

EC=1−tan 25 FC=1−tan 20 tan ∠FEC=((FC)/(EC))=((1−tan 20)/(1−tan 25)) x=180−65−tan^(−1) ((1−tan 20)/(1−tan 25))=65°

$${EC}=\mathrm{1}−\mathrm{tan}\:\mathrm{25} \\ $$$${FC}=\mathrm{1}−\mathrm{tan}\:\mathrm{20} \\ $$$$\mathrm{tan}\:\angle{FEC}=\frac{{FC}}{{EC}}=\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{20}}{\mathrm{1}−\mathrm{tan}\:\mathrm{25}} \\ $$$${x}=\mathrm{180}−\mathrm{65}−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}−\mathrm{tan}\:\mathrm{20}}{\mathrm{1}−\mathrm{tan}\:\mathrm{25}}=\mathrm{65}° \\ $$

Commented by I want to learn more last updated on 07/May/20

Thanks sir. I really appreciate

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate} \\ $$

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