Question Number 195344 by Mingma last updated on 31/Jul/23
Answered by kapoorshah last updated on 31/Jul/23
$${cos}\:\alpha\:=\:{cos}\:\alpha \\ $$$$\frac{{a}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{a}.\mathrm{6}}\:=\:\frac{{b}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{b}.\mathrm{6}} \\ $$$$\:\frac{{a}^{\mathrm{2}} \:+\:\mathrm{20}}{{a}}\:=\:\frac{{b}^{\mathrm{2}} \:+\:\mathrm{20}}{{b}} \\ $$$${a}^{\mathrm{2}} {b}\:+\:\mathrm{20}{b}\:=\:{ab}^{\mathrm{2}\:} +\:\mathrm{20}{a} \\ $$$$\left({a}\:−\:{b}\right)\left({ab}\:−\:\mathrm{20}\right)\:=\:\mathrm{0} \\ $$$${a}\:=\:{b}\:\:\:\left({rejected}\right) \\ $$$${ab}\:=\:\mathrm{20} \\ $$
Commented by kapoorshah last updated on 31/Jul/23
Commented by Mingma last updated on 31/Jul/23
Perfect ��
Answered by HeferH last updated on 31/Jul/23
$${by}\:{drawing}\:{two}\:{heights}: \\ $$$$\mathrm{36}\:−\:\left({a}\:+\:\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{16}\:−\:\left(\frac{{b}−{a}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{20}\:=\:{a}^{\mathrm{2}} \:+\:{a}\left({b}−{a}\right) \\ $$$$\:\mathrm{20}\:=\:{ab}\: \\ $$