Question

plzz answer this ....​

Answer 1

Answer:

x=50°

Step-by-step explanation:

angle T = 70° (opposite angle of parallelogram TINS)

angle J + U =180° (adjacent sides of parallelogram)

J+120=180

J=70°

now, in triangle TJX-

angle T + angle J + angle X = 180°( by angle sum property of a triangle)

solving, we get-

70°+60°+X=180°

130°+X=180°

X=180°-130°

X=50°.

hope this helps.

pls mark as brainliest

Answer 2

Answer:

Step-by-step explanation: In llgm RAJU,

∠RUJ= ∠JAR (Opp. angles of llgm are equal)

∴ ∠JAR= 120°

∠JAI+ ∠JAR= 180° (Linear Pair)

⇒ ∠JAI + 120°= 180°

∴∠JAI= 60°

Now in llgm TINS,

∠SNI=∠TIA (corresponding angles)

∴ ∠TIA= 70°

Now, in triangle OAI,

∠TIA +∠JAI + ∠AOI= 180°

⇒ 70°+ 60°+ ∠AOI= 180°

⇒ ∠AOI =50°

Now ∠AOI= ∠x (vertically opp. angles)

∴ ∠x= 50°