Is the answer $10$ ?

yes i got the same

extend line $C$ to meet $AB$ at $D$ making line $C$ as transversal $ \angle ADC = 130^{o}$ USING EXTERIOR ANGLE PROPERTY $\angle DCA + \angle CAD = 130^{o}$ $ \angle DCA = 180^{o} - 60^{o}$ = $120^{o}$ THEREFORE $ \angle ABC = 10^{o}$

This was RMO???

RishiNandha_M wrote: This was RMO??? Actually its confusing. It is between PRMO and RMO

Let the set square vertices be CMN where angle M= 90 extend MC too meet AB at K then angle BKC = 50 degree since the ruler opposite sides are parallel angle KCB = 120 (exterior angle property) so in triange KCB -> angle ABC = angle KBC = 10 by angle sum property

pentagon ABC(intersection of line c and the ruler top side) total angle is 540, subtract 180+50+300, and you get 10 as the final answer