Question

In figure given below abcd is a kite and diagonals intersect at o. If angle dab=112 and angle dcb= 64 . Find angle odd and angle oba

Answer 1

Answer:

∠ODC = 58°    and  ∠OBA = 34°  

Step-by-step explanation:

We have, ABCD as the given kite. We know that in a kite AD = AB , BC = DC.Also it is given that ∠DAB = 112°; ∠BCD = 64° and diagonals AC and BD intersect each other at O.

Now, in ΔADC and ΔABC, we have

AD = AB    (given)

DC = BC   (given)

AC = AC    (Common)

ΔADC ≅ ΔABC  (SSS congruency criterion)

⇒∠OAD = ∠OAB   (CPCT)   .....(1)

⇒∠OCD = ∠OCB   (CPCT)   ......(2)

Now, ∠DAB = 112°⇒∠OAD + ∠OAB

                   = 112°⇒2∠OAD = 112°     [using (1)]

                   ⇒∠OAD =56° =∠OAB      

           Now,   ∠BCD = 64°⇒∠OCD + ∠OCB = 64°⇒2∠OCD = 64°    [using(2)]⇒

∠OCD = 32° = ∠OCB

We know that diagonals of a kite intersect each other at right angle, so ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Now, in ΔDOC, we have∠ODC + ∠COD + ∠OCD = 180°    [Angle sum property of triangle]

⇒∠ODC + 90° + 32° = 180°⇒∠ODC = 180° − (90° + 32°)

= 180° − 122° = 58°

In ΔAOB, we have∠OBA + ∠AOB + ∠OAB = 180°    [Angle sum property of triangle]

⇒∠OBA +  90° + 56° = 180°⇒∠OBA

= 180° − (90°+56°) = 34°

So, ∠ODC = 58°    and  ∠OBA = 34°  

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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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