the given figure is a rectangle share diagonals bisect at O. If AOB is 108 degree then find
I)ABO
II)ADO
III)OCB
Answers
Answer:
ABO=36 °
\angle ADO = 54∠ADO=54 °
\angle OCB = 54∠OCB=54 °
Step-by-step explanation:
In rectangle ABCD , diagonals intersect at point O
AC = BD (diagonals of a rectangle are equal)
\frac{1}{2} AC = \frac{1}{2} BD
2
1
AC=
2
1
BD
in triangle AOB (isoceles triangle)
let the angle OBA = x
\angle AOB +\angle OAB + \angle OBA = 180∠AOB+∠OAB+∠OBA=180
108 +x+x = 180
2x = 180-108
2x = 72
x = 36
\angle OBA = \angle OAB = 36∠OBA=∠OAB=36
i.e \angle ABO = 36∠ABO=36 °
\angle ABO = \angle ODC = 36∠ABO=∠ODC=36
We know that The four interior and exterior angles are 90 deg.
i.e
\begin{gathered}\angle ODC + \angle AD0 = 90\\\\36 +\angle ADO = 90\\\angle ADO = 90- 36 = 54\end{gathered}
∠ODC+∠AD0=90
36+∠ADO=90
∠ADO=90−36=54
i.e \angle ADO = 54∠ADO=54 °
similarly ,
\begin{gathered}\angle OAB = \angle OCD = 36\\\angle OCB +\angle OCD = 90\\\angle OCB + 36 = 90\\\angle OCB = 90-36\\\angle OCB = 54\end{gathered}
∠OAB=∠OCD=36
∠OCB+∠OCD=90
∠OCB+36=90
∠OCB=90−36
∠OCB=54
hence ,
\angle ABO = 36∠ABO=36 °
\angle ADO = 54∠ADO=54 °
\angle OCB = 54∠OCB=54 °
Diagonals of rectangle ABCD intersect at O.If angle AOB = 30°, find angle COD angle OCD.
Answer:
From the figure it is given that, ABCD is a rectangle and diagonals intersect at O
∠AOB=118∘
(i) Consider the ΔAOB
∠OAB=∠OBA
Let us assume ∠OAB=∠OBA=y∘
We know that, sum of measures of interior angles of triangle is equal to 180∘.
∠OAB+∠OBA+∠AOB=180∘
y+y+118∘
=180∘
2x+118∘=180∘
By transposing we get, 2y=180∘−118∘
2y=62∘
y=62/2
y=31∘
So,∠OAB=∠OBA=31∘
Therefore, ∠ABO=31∘
(ii) We know that sum of liner pair angles is equal to 180∘.
∠AOB+∠AOD=180∘
118∘+∠AOD=180∘
∠AOD=180∘−118∘
∠AOD=62∘
Now consider the △AOD
Let us assume the ∠ADO=∠DAO=x
∠AOD+∠ADO+∠DAO=180∘
62∘+x+x=180∘
62∘+2x=180∘
By transposing we get, 2x=180∘−622x=118∘
x=118∘/2
x=59∘
Therefore, ∠ADO=59∘
(iii) ∠OCB=∠OAD=59∘
... [because alternate angles are equal]
Hope this helps u