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(a) 32° (exterior angle=sum of opp. int. angles, base angles of isosceles triangle)

(b) x°= 32+90=122° (alternate angles, PQ//RS)

(c) since y+50= 122 (exterior angle=sum of opp. int. angles in triangle), y°=72° #

(d) angle HGQ= 58° (angles on straight line),

z°= 58° (corresponding angles, PD//FH)

(b) x°= 32+90=122° (alternate angles, PQ//RS)

(c) since y+50= 122 (exterior angle=sum of opp. int. angles in triangle), y°=72° #

(d) angle HGQ= 58° (angles on straight line),

z°= 58° (corresponding angles, PD//FH)

Date Posted:
8 months ago