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Question-92518




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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