Question

The diagonals of rhombus ABCD intersect at the point ‘O’. If ∠BDC=50° , then ∠OAB is​

Answer 1

Answer:

40°

Step-by-step explanation:

     Its given ABCD is a IIgram and AC and BD are its diagonals intersecting at point O.

          Given : angle BOC = 900 

                            angle BDC = 500 

          To find : angle OAB 

          Answer : i) angle BOC + angle COD = 180° (Linear Pair)

                           angle COD = 180° - angle BOC

                           angle COD = 180° - 90°  

                           angle COD = 90°

                       ii) angle COD + angle ODC + angle OCD = 180 (Angle Sum Property)

                           angle OCD = 180 - 90 - 50

                           angle OCD = 40°

                       iii) angle OCD = angle OAB (Alternate Interior Angles)

                            so, therefore angle OAB = 40°.  

Answer 2

Given:

ABCD is a rhombus, its diagonals intersect at 'O'.

∠BCD = 50°

To Find:

∠OAB

Solution:

 ∠BOC and ∠COD are linear pairs.

⇒ ∠BOC + ∠COD = 180°

⇒ 90° + ∠COD = 180°

⇒ ∠COD = 180°- 90°

⇒ ∠COD = 90°

Now,

 ∠COD + ∠BDC + ∠OCD = 180° [angle sum property of a triangle]

⇒ 90° + 50° + ∠OCD = 180°

⇒ ∠OCD = 180° - 140°

⇒ ∠OCD = 40°

Then, ∠OCD = ∠OAB ( Alternate interior angles are equal)

So, ∠OAB = 40°

Therefore, ∠OAB = 40°