Question

in a parallelogram any possibilty for a quadrilateral ABCD, ∠A=65⁰degree ​

Answer 1

Hence ∠A=∠C

⇒3y=3x+3 ...(1)

and ∠B=∠D

∴ ∠D=2y−5 ...(2)

We know that the sum of angles of a parallelogram is 360

∘

⇒(3y)+(3x+3)+(2y−5)+(2y−5)=360

⇒3y+3y+4y−10=360

⇒10y−10=360

⇒10y=360+10

⇒10y=370

⇒y=

10

370

=37

∘

Substitute y=37

∘

in (1)

⇒3×(37

∘

)=3x+3

∘

⇒111=3x+3

⇒3x=111−3

⇒3x=108

⇒x=

3

108

=36 ∘ (Ans)