Question Number 186492 by mr W last updated on 05/Feb/23
Commented by mr W last updated on 05/Feb/23
$${unsolved}\:{question}\:{in}\:#\mathrm{186283} \\ $$
Answered by mr W last updated on 05/Feb/23
Commented by mr W last updated on 05/Feb/23
$${the}\:{diagram}\:{given}\:{in}\:{the}\:{question}\:{is} \\ $$$${drawn}\:{not}\:{to}\:{scale}\:{and}\:{the}\:{lengthes} \\ $$$${given}\:{are}\:{not}\:{properly}\:{selected},\: \\ $$$${therefore}\:{the}\:{diagram}\:{is}\:{in}\:{reality} \\ $$$${not}\:{possible}.\:{but}\:{the}\:{method}\:{for}\:{the} \\ $$$${solution}\:{is}\:{correct}. \\ $$
Commented by mr W last updated on 05/Feb/23
Commented by mr W last updated on 05/Feb/23
$${with}\:{these}\:{values}\:{we}\:{get} \\ $$$$\frac{\mathrm{6}}{\mathrm{6}+{x}}=\frac{\mathrm{7}−{y}}{\mathrm{7}+\mathrm{3}}=\frac{{y}}{\mathrm{12}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{42}}{\mathrm{11}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{90}}{\mathrm{7}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{12}^{\mathrm{2}} +\left(\mathrm{6}+\frac{\mathrm{90}}{\mathrm{7}}\right)^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }{\mathrm{2}×\mathrm{12}×\left(\mathrm{6}+\frac{\mathrm{90}}{\mathrm{7}}\right)}=\frac{\mathrm{445}}{\mathrm{504}} \\ $$$$\Rightarrow\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{445}}{\mathrm{504}}\approx\mathrm{28}° \\ $$
Commented by mr W last updated on 05/Feb/23
