Diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90º and ∠BDC = 50º, then ∠OAB is
a. 90º
b. 50º
c. 40º
d. 10º
Solution:
We know that
ABCD is a parallelogram with AC and BC as diagonals that intersect at O
Given,
∠BOC = 90º
∠BDC = 50º
We have to determine ∠OAB
∠BOC + ∠COD = 180º [Linear Pair]
By substituting the values
90º + ∠COD = 180º
∠COD = 180º - 90º
∠COD = 90º
As O lies on BD
∠ODC = ∠BDC
∠ODC = 50º
From the angle sum property of a triangle
∠ODC + ∠COD + ∠OCD = 180º
Substituting the values
50º + 90º + ∠OCD = 180º
So we get
∠OCD = 180º - 140º
∠OCD = 40º
As O lies on AC
∠ACD = ∠OCD
∠ACD = 40º
Here DC || AB
So we get
∠CAB = ∠ACD
∠OAB = 40º
Therefore, ∠OAB = 40º.
✦ Try This: Diagonals of a parallelogram PQRS intersect at O. If ∠QOR = 60º and ∠QSR = 40º, then ∠OPQ is a. 90º, b. 50º, c. 40º, d. 10º
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8
NCERT Exemplar Class 9 Maths Exercise 8.1 Sample Problem 1
Diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90º and ∠BDC = 50º, then ∠OAB is , a. 90º, b. 50º, c. 40º, d. 10º
Summary:
Diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90º and ∠BDC = 50º, then ∠OAB is 40º
☛ Related Questions:
- Three angles of a quadrilateral are 75º, 90º and 75º. The fourth angle is a. 90º, b. 95º, c. 105º, d . . . .
- A diagonal of a rectangle is inclined to one side of the rectangle at 25º. The acute angle between t . . . .
- ABCD is a rhombus such that ∠ACB = 40º. Then ∠ADB is a. 40º, b. 45º, c. 50º, d. 60º