Question

ABCD is a rhombus, if angle ACB=50 then find angle ADB

Answer 1

Answer:

Step-by-step explanation:

AB=BC(In a rhombus all sides are equal)

AngleACB=BAC=50

Angle ACB=AngleCAD(alternate angles)

AngleA=AngleBAC+AngleCAD

=50+50

=100

In triangleBAD

AB=AD

AngleADB=AngleABD

BAD+ADB+ABD=180

100+ADB+ADB=180

2ADB=180_100

ADB=80/2

=40

Answer 2

Answer:

The Angle of ADB is 40°

Step-by-step explanation:  

• In context to the given question we have to find the ∠ADB in rhombus ABCD.
• GIVEN:

1. ∠ACB = 50°
2. As it is rhombus ; diagonals will bisect at 90°.

⇒ Therefore, let the center be O

∠AOD= ∠AOB= ∠BOC= ∠COD= 90°

Now, In Δ BOC

⇒ ∠OCB +∠COB+∠OBC =180°  ( angle sum property of a triangle)

⇒ 50° + 90° + ∠OBC = 180°

⇒ ∠OBC = 180° - 140°

⇒ ∠OBC = 40°

Now, as we know , ABCD is a rhombus and rhombus is a parallelogram

∴ AD║BC

⇒  ∠OBC = ∠ADB   ( alternate interior angle property )

∴  ∠ADB = 40°

The Angle of ADB is 40°