Question

in the adjoining figure, ABCD is a trapezium, if AOB = 126
and PDC = QCD = 52°, then find x & y.
(A) x = 52°, y = 52°
(C) x = 101° y = 25°
(B) x = 25y = 101°
(D) x = 52, y = 101°​

Answer 1

Solution :-

Construction :- Produce AP and BQ to meet at R.

In ∆RDC we have,

→ ∠RDC = ∠RCD (given 52°) ---------- Eqn.(1)

so,

→ DR = CR (sides opposite to equal angles are equal in length.)

now,

→ ∠RAB = ∠RDC (given that, ABCD is a trapezium. so, AB || DC , therefore, corresponding angles are equal). -------- Eqn.(2)

Similarly,

→ ∠RBA = ∠RCD (corresponding angles .) -------- Eqn.(3)

from Eqn.(1) , Eqn.(2) and Eqn.(3) we can conclude that,

→ ∠RAB = ∠RBA

or,

→ AR = RB .

then,

→ AD = AR - DR = RB - CR = BC

therefore, we can conclude that, ABCD is an isosceles trapezium.

hence,

→ OA = OB

→ ∠OAB = ∠OBA.

now, in ∆AOB we have,

→ ∠OAB + ∠OBA + ∠AOB = 180° (By angle sum property.)

→ 2∠OAB + 126° = 180°

→ 2∠OAB = 180° - 126°

→ 2∠OAB = 54°

→ ∠OAB = 27°

therefore,

→ ∠x = ∠DAB - ∠OAB

→ ∠x = 52° - 27°

→ ∠x = 25° (Ans.)

also, in ∆ABC, we have,

→ ∠ACB + ∠CAB + ∠ABC = 180° (By angle sum property.)

→ y + 27° + 52° = 180°

→ y = 180° - 27° - 52°

→ y = 180° - 79°

→ y = 101° (Ans.)

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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