Question

in the given figure ABCD is a rectangle whose diagonals intersect at O diagonal AC is produced to E and angle is CD is equal to 140 degree find the angles of triangle OAB

Answer 1

50 THE ANSWER RIGHT.....................

Answer 2

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

= 90o + 60o

= 150o … (iv)

Also, EC = CD [From (iii)]

So, ∠DEC = ∠CDE … (v)

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

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

2 ∠DEC = 180o – 150o = 30o

∠DEC = 30o/2

∠DEC = 15o … (vi)

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

∠BED + ∠DEC = 60o

xo + 15o = 60o [From (vi)]

x = 60o – 15o

x = 45o

Hence, the value of x is 45o.

(b) Given, ABCD is a rectangle

∠ECD = 146o

As ACE is a straight line, we have

146o + ∠ACD = 180o [Linear pair]

∠ACD = 180o – 146o = 34o … (i)

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

From (i) and (ii), we have

∠CAB = 34o ⇒ ∠OAB = 34o … (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 = 34o … (v)

Now,

∠AOB + ∠OBA + ∠OAB = 180o

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

∠AOB + 68o = 180o

∠AOB = 180o – 68o = 112o

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

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

Let ∠OAB = 2xo

Then,

∠OBA = 2xo

We know that diagonals of rhombus intersect at right angle,

So, ∠OAB = 90o

Now, in ∆AOB

∠OAB + ∠OBA = 180o

90o + 3xo + 2xo = 180o

90o + 5xo = 180o

5xo = 180o – 90o = 90o

xo = 90o/5 = 18o

Hence,

∠OAB = 3xo = 3 x 18o = 54o

OBA = 2xo = 2 x 18o = 36o and

∠AOB = 90o

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