In the figure given below, ABCD is a rectangle
Find angles of AOD
Answers
Answer:
ABO = BAO = 30 and AOB = 120
Step-by-step explanation:
Angle ABO = 180 - 150 = 30 ( exterior angle property)
angle ABO will be equal to angle BAO ( the diagonals of a rectangle are equal in length)
Angle AOB will be 180 - 30 - 30 = 120. (sum of angles in a triangle is 180)
So,ABO = BAO = 30 and AOB = 120
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