Question Number 165531 by ajfour last updated on 03/Feb/22
Commented by ajfour last updated on 03/Feb/22
$$\:\:\:\:\:\:\:\:\:\:{For}\:{class}\:{XI},\:{perhaps}! \\ $$
Answered by mr W last updated on 03/Feb/22
Commented by mr W last updated on 03/Feb/22
$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}}{{a}} \\ $$$${FC}={b}\left(\mathrm{sin}\:\alpha+\frac{\mathrm{1}}{\mathrm{cos}\:\alpha}\right)={a}\:\mathrm{cos}\:\alpha \\ $$$$\frac{{b}}{{a}}=\frac{\mathrm{cos}^{\mathrm{2}} \:\alpha}{\mathrm{1}+\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}=\frac{{a}^{\mathrm{2}} }{\left(\mathrm{1}+{a}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{a}}{\mathrm{1}+{a}^{\mathrm{2}} }\right)} \\ $$$${b}=\frac{{a}^{\mathrm{3}} }{{a}^{\mathrm{2}} +{a}+\mathrm{1}} \\ $$
Commented by ajfour last updated on 03/Feb/22
$${very}\:{pleasant},\:{Sir}! \\ $$