Question

These 2 are sought please any body say the answer
Pleaseeeeeeee

Answer 1

27. In ∆ABC due to angle sum property,

∠BAC + ∠ABC + ∠ACB = 180°

As AB = AC so angle opposite to equal sides are also equal.

100° + ∠ACB + ∠ACB = 180° [ :. AB = AC ]

2∠ACB = 80°

∠ACB = 40°

Now,

∠ACB + ∠ACE + ∠ECD = 180° [ Linear Pair ]

40° + ∠ACE + 40° = 180°

∠ACE = 100°

But ∠BAC = 100°, so

∠ACE = ∠BAC

When alternate angles are equal then two lines are parallel

⇒Hence AB || CE

28. In rectangle PQRS, equal diagonals bisect each other.

PR = QS

½ PR = ½ QS

OQ = OR

So, ∠OQR = ∠ORQ [ Angle opp to equal sides are also equal ]

∠OQR = 30°

But, ∠PQR = 90°

90° = ∠PQS + 30°

⇒60° = ∠PQS

Answer 2

27.) Given, AB = AC and triangle ABC is an isosceles triangle.

Let the angle be B and C be x°.

Sum of interior angles of a triangle = 180°

100° + x° + x° = 180°

2x° = 180° - 100°

x° = 80°/2 = 40°

So, B and C = 40°

BCD forms linear pair.

So, BCA + ACE + ECD = 180°

40° + ACE + 40° = 180°

ACE = 180° - 80°

Angle ACE = 100°

Hence, Angle BAC = Angle ACE

Alternate angles are equal.

So, AB // CE.



28.) PQ //RS and PS // QR

Angle OQR = Angle ORQ = 30° [ OQ = OR ]

Each Angle of rectangle is 90°.

SO, Angle PQO + Angle OQR = 90°

Angle PQO = 90° - 30°

$<h3>Angle PQO = 60°</h3>$