Question

Fie unghiurile AOB si BOC adiacente astfel incat m(BOC)= 3 supra 4 inmultit cu m(AOB) , (OM bisectoarea unghiului AOB si (ON astfel incat m(MON)=90 grade.Aflati masurile unghiurilor AOB si BOC stiind ca m(CON)=20 grade.Va rog ajutati-ma chiar vreau sa stiu rezolvarea.Multumesc frumos.

Answer 1

$\angle AOB \ ; \angle BOC -adiacente \\\\ mas \ \angle AOB=x \\\\ mas \ \angle BOC= \frac{3}{4}* mas \ \angle AOB= \frac{3}{4}*x \to \frac{3x}{4} \\\\\\\ [ OM- \ bisect. \ \angle AOB \implies \\\\ \implies mas \ \angle AOM=mas \ \angle MOB= \frac{mas \ \angle AOM}{2} =\frac{x}{2}$  

$\underline{mas \ \angle CON=20^0} \\\\\\ mas \ \angle MOB+mas \ \angle BON=90^0 \\\\ \frac{x}{2}+mas \ \angle BON=90^0 \\\\ mas \ \angle BON= 90^0-\frac{x}{2} \\\\\\ mas \ \angle CON+mas \ \angle BON=\frac{3x}{4} \\\\ 20^0+mas \ \angle BON=\frac{3x}{4} \\\\ mas \ \angle BON= \frac{3x}{4}-20^0$

$\frac{3x}{4}-20^0=90^0-\frac{x}{2} \\\\ -20^0-90^0=-\frac{x}{2}^{(2}-\frac{3x}{4} \\\\ -110^0= \frac{-2x-3x}{4} \\\\ -440^0=-5x \\\\ x=88^0 \\\\\\ \boxed{mas \ \angle AOB=88^0} \\\\\\ mas \ \angle BOC= \frac{3x}{4} =\frac{3*\not 88^0}{\not 4}= 3*22^0 \\\\\\ \boxed{mas \ \angle BOC=66^0}$