Question

ABCD is a rhombus, diagonals intersect at O. Angle OAB is 36°. Find angle ODA.
Answer As Soon As Possible

Answer 1

A

90

ABCD is a rhombus.

AC and BD are diagonals of rhombus.

We know that, diagonals of a rhombus bisect each other perpendicularly .

∴ AC⊥BD

i.e.AO⊥BO

∴ ∠AOB=90

o

Answer 2

Step-by-step explanation:

Since, EBC is an equilateral triangle, we have

EB = BC = EC … (i)

Also, ABCD is a square

So, AB = BC = CD = AD … (ii)

From (i) and (ii), we get

EB = EC = AB = BC = CD = AD … (iii)

Now, in ∆ECD

∠ECD = ∠BCD + ∠ECB

= 90° + 60°

= 150°… (iv)

Also, EC = CD [From (iii)]

So, ∠DEC = ∠CDE … (v)

∠ECD + ∠DEC + ∠CDE = 180° [Angles sum property of a triangle]

150° + ∠DEC + ∠DEC = 180° [Using (iv) and (v)]

2 ∠DEC = 180°– 150° = 30°

∠DEC = 30°/2

∠DEC = 15° … (vi)

Now, ∠BEC = 60° [BEC is an equilateral triangle]

∠BED + ∠DEC = 60°

x°+ 15° = 60° [From (vi)]

x = 60° – 15°

x = 45°

Hence, the value of x is 45°.

(b) Given, ABCD is a rectangle

∠ECD = 146°

As ACE is a straight line, we have

146o + ∠ACD = 180° [Linear pair]

∠ACD = 180° – 146° = 34° … (i)

And, ∠CAB = ∠ACD [Alternate angles] … (ii)

From (i) and (ii), we have

∠CAB = 34° ⇒ ∠OAB = 34° … (iii)

In ∆AOB

AO = OB [Diagonals of a rectangle are equal and bisect each other]

∠OAB = ∠OBA … (iv) [Equal sides have equal angles opposite to them]

From (iii) and (iv),

∠OBA = 34° … (v)

Now,

∠AOB + ∠OBA + ∠OAB = 180°

∠AOB + 34° + 34° = 180° [Using (3) and (5)]

∠AOB + 68° = 180°

∠AOB = 180° – 68° = 112°

Hence, ∠AOB = 112°, ∠OAB = 34° and ∠OBA = 34°

(c) Here, ABCD is a rhombus and diagonals intersect at O and ∠OAB : ∠OBA = 3 : 2

Let ∠OAB = 2x°

Then,

∠OBA = 2x°

We know that diagonals of rhombus intersect at right angle,

So, ∠OAB = 90°

Now, in ∆AOB

∠OAB + ∠OBA = 180°

90° + 3x° + 2x° = 180°

90° + 5x°= 180°

5x°= 180° – 90° = 90°

x° = 90°/5 = 18°

Hence,

∠OAB = 3x° = 3 x 18° = 54°

OBA = 2x° = 2 x 18° = 36° and

∠AOB = 90°