# 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

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