Question Number 184638 by HeferH last updated on 09/Jan/23
Commented by HeferH last updated on 09/Jan/23
$${find}\:``{x}''\:{if}\:{the}\:{base}\:\left(\mathrm{60}°\:-\:\mathrm{2}\alpha\right)\:{has}\:{a}\:{length}\:{of}\:\mathrm{12} \\ $$
Commented by mr W last updated on 09/Jan/23
$${can}\:{you}\:{exactly}\:{specify}\:{what}\:{is}\:\mathrm{6},\:\mathrm{4} \\ $$$${and}\:{x}?\:{like}\:{this}\:{way}: \\ $$
Commented by mr W last updated on 09/Jan/23
Commented by HeferH last updated on 09/Jan/23
$${I}\:{forgot}\:{but}\:{next}\:{time}\:{I}'{ll}\:{use}\:{more}\:{colors} \\ $$$$\left.\:\left({Your}\:{drawing}\:{is}\:{correct}\right)\::\right) \\ $$
Commented by mr W last updated on 09/Jan/23
$${then}\:{something}\:{is}\:{wrong}! \\ $$$$\alpha+\beta=\mathrm{60}° \\ $$$${say}\:{y}={right}\:{side} \\ $$$$\frac{{y}}{\mathrm{sin}\:\mathrm{60}}=\frac{\mathrm{12}}{\mathrm{sin}\:\mathrm{2}\beta} \\ $$$$\frac{{y}}{\mathrm{sin}\:\left(\alpha+\beta\right)}=\frac{\mathrm{6}}{\mathrm{sin}\:\beta} \\ $$$$\frac{\mathrm{6}}{\mathrm{sin}\:\beta}=\frac{\mathrm{12}}{\mathrm{sin}\:\mathrm{2}\beta}=\frac{\mathrm{12}}{\mathrm{2}\:\mathrm{sin}\:\beta\:\mathrm{cos}\:\beta} \\ $$$$\Rightarrow\mathrm{cos}\:\beta=\mathrm{1}\:\Rightarrow\beta=\mathrm{0}°\:!! \\ $$
Commented by HeferH last updated on 10/Jan/23
$$\:{The}\:{problem}\:{says}: \\ $$$$\:``{In}\:{a}\:{triangle}\:{ABC}\:{draw}\:{the}\:{interior}\:{bisectors} \\ $$$$\:{BM}\:{and}\:{CN}\:{which}\:{intersect}\:{at}\:{P}.\: \\ $$$$\:{if}\:{the}\:\:{angle}\:{BAC}\:=\:\mathrm{60}°,\:{AC}\:=\:\mathrm{12},\:{PM}\:=\:\mathrm{4}\:{and} \\ $$$$\:{PC}\:=\:\mathrm{6},\:{find}\:{AN}.\:'' \\ $$$$\:{then}\:{maybe}\:{my}\:{drawing}\:{its}\:{no}\:{accurate}\::/ \\ $$