Question

ABCD is rectangle in given figure below. If ∠AOB = 118°, Calculate: i) ∠ABO ii) ∠ADO iii) ∠OCB​

Answer 1

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

∘

−62

2x=118

∘

x=118

∘

/2

x=59

∘

Therefore, ∠ADO=59

∘

(iii) ∠OCB=∠OAD=59

∘

... [because alternate angles are equal]

Step-by-step explanation:

❥︎hey dear here is ur answer

❥︎plz like my answer and mark me as brilliant

❥︎bye tc have a great day♡︎

Answer 2

→ Hey Mate,

→ Given Question:-

→ ABCD is a rectangle in the given figure below. If ∠AOB = 118°, Calculate: i) ∠ABO ii) ∠ADO iii) ∠OCB

----------------------------------------------------

→ Solution:-

→ From the given figure it is given that, ABCD is a rectangle & diagonals

→ intersect at o.

→ ∠AOB = 118°

----------------------------------------------------

→ (1$^s^t$ ) Solution:-

→ Consider the Δ AOB

→ ∠OAB = ∠OBA

→ Take ∠OAB = ∠OBA = y°

→ The sum of measures of interior angles of a triangle is equal to 180°.

→ ∠OAB + ∠OBA + ∠AOB = 180°

→ y + y + 118° = 180°

→ 2x + 118° = 180°

→ We get , 2y = 180° - 118° [ ∵ By transposing ]

→ 2y = 62°

→ y = $\frac{62}{2}$

→ y = 31°

→ So , ∠OAB = ∠OBA = 31°

→ ∴ , ∠ABO = 31°

----------------------------------------------------

→ ( 2nd)  solution:-

→ The Sum of Linear pair angles is equal to 180°

→ ∠AOB + ∠AOD = 180°

→ 118° + ∠AOD = 180°

→ ∠AOD = 180° - 118°

→ ∠AOD = 62°

→ Consider the ΔAOD

→ Let us take  the ∠ADO = ∠DAO = x

→ ∠AOD + ∠ADO + DAO = 180°

→ 62° + X + X =180°

→ 62° +  2X    = 180°

→ We get , 2x = 180° - 62  [ ∵ By transposing ]

→ 2x = 118°

→ x = $\frac{118}{y}$ degree

→ x = 59°

→ ∴ , ∠ADO = 59°

----------------------------------------------------

→ (3rd) solution:-

→ ∠OCB = ∠OAD = 59° ... [∵ Alternate angles are equal]

----------------------------------------------------

→ Similar Question:-

→ In the given figure, MNOQ is a rectangle. if ∠MSN = 118° , calculate

→ 1) ∠MNS

→ 2) ∠MQS

→ 3) ∠SON

→( Hint :- Attachment in the figure .)

--------------------------------------------------------------------------------------------------------