The Total solution for NCERT class 6-12
In the given figure, if PQтКе PS, PQ|| SR, тИаSQR = 28┬║and тИаQRT =65┬║, then find the values of┬аx┬аand┬аy.
It is given that PQ || SRand QR is a transversal line.
тИаPQR = тИаQRT (Alternate interior angles)
x┬а+ 28┬║ = 65┬║
x┬а= 65┬║ тИТ 28┬║
x┬а= 37┬║
By usingthe angle sum property for ╬ФSPQ, we obtain
тИаSPQ +┬аx┬а+┬аy┬а= 180┬║
90┬║ + 37┬║+┬аy┬а= 180┬║
y┬а= 180┬║ тИТ 127┬║
y┬а= 53┬║