Question

In the given figure, ABCD is a rectangle and diagonals intersect at O. If ∠AOB=118∘, find (i) ∠ABO (ii) ∠ADO (iii) ∠OCB​

Answer 1

Answer:

(i) 31°

(ii) 59°

(iii) 59°.

Step-by-step explanation:

Given,

• ABCD is a rectangle.
• ∠AOB is 118°.

In ∆ABD and ∆ABC.

⇒AB = AB [Common side]

⇒∠A = ∠B = 90° [All angles of rectangle are right angled]

⇒ AD = BC [Opposite sides of rectangle are equal]

By SAS congruence rule :

∆ABD ≌ ∆ABC .

By CPCT :

→ ∠ABO = ∠OAB

→ ∠ADO = ∠OCB

Now finding angles :

(i) ∠ABO

In ∆AOB :

By angle sum property of triangle :

$\longrightarrow$ ∠AOB + ∠ABO + ∠OAB = 180°

$\longrightarrow$ 118° + ∠ABO + ∠ABO = 180° [∠AOB = 118° and ∠ABO = ∠OAB as we proved above]

$\longrightarrow$ 118° + 2∠ABO = 180°

$\longrightarrow$ 2∠ABO = 180° - 118°

$\longrightarrow$ 2∠ABO = 62°

$\longrightarrow$ ∠ABO = 62°/2

$\longrightarrow$ ∠ABO = 31°

Thus,

∠ABO is 31°.

(ii) ∠ADO

In ∆ABD :

By angle sum property :

$\longrightarrow$ ∠ADO + ∠BAD + ∠ABD = 180°

$\longrightarrow$ ∠ADO + 90° + 31° = 180° [∠BAD = 90° as it is one of angle of rectangle and ∠ABD or ∠ABO = 31°]

$\longrightarrow$ ∠ADO + 121° = 180°

$\longrightarrow$ ∠ADO = 180° - 121°

$\longrightarrow$ ∠ADO = 59°

Thus,

∠ADO is 59°.

(iii) ∠OCB

→ ∠OCB = ∠ADO [As we proved above]

→ ∠ADO = 59° [As we have find it in (ii) part]

So,

→ ∠OCB = 59°

Thus,

∠OCB is 59°.

Answer 2

Answer:

(i) 31°

(ii) 59°

(iii) 59°.

Step-by-step explanation:

$\tt{Given : }$

• ABCD is a rectangle.
• ∠AOB is 118°.

Also, In ∆ABD and ∆ABC.

$\implies{ \sf{AB = AB (Common \: side)}}$

$\sf{ \implies \: \angle \: A = \angle \: B = 90° (All \: angles \: of \: rectangle \: are \: right \: angles)}$

$\sf{ \implies{AD = BC \: (Opposite \: sides \: of \: rectangle \: are \: equal)}}$

By applying SAS (side angle side) congruence rule :

$\boxed{ \tt{ \triangle \: ABD \cong \triangle \: ABC }}$

By using CPCT (corresponding sides of the corresponding triangle) :

➺ ∠ABO = ∠OAB

➺ ∠ADO = ∠OCB

Now, According to the question let's find angles :

(i) ∠ABO

In ∆AOB :

$\sf{Using \: angle \: sum \: property \: of \: triangle :}$

★ Angle sum property of a ∆ states that sum of all the angles of a triangle is 180°.

$\dashrightarrow \tt{ \angle \: AOB + \angle \: ABO + \angle \: OAB = 180 {}^{ \circ} }$

${\tt{ \dashrightarrow \: 118 {}^{ \circ} + \angle \: ABO + \angle \: ABO = 180 {}^{ \circ} \{\angle \: AOB = 118 {}^{ \circ} \: and \: \angle \: ABO = ∠OAB \: as \: we \: have \: proved \: above \}}}$

${\tt{\dashrightarrow \: 118 {}^{ \circ} + 2 \angle \: ABO = {180}^{ \circ} }}$

$\tt{\dashrightarrow \: 2 \angle \: ABO = 180^{ \circ} - 118^{ \circ} }$

$\tt{ \dashrightarrow \: 2 \angle \: ABO = {62}^{ \circ} }$

$\tt{ \dashrightarrow \: \angle \: ABO = \cancel{\dfrac{62 {}^{ \circ} }{2 {}^{}}}}$

$\tt{\dashrightarrow{ \angle \: ABO = 31^{ \circ} }}$

Hence,

➸ ∠ABO is 31°.

(ii) ∠ADO

In ∆ABD :

$\sf{By \: using \: angle \: sum \: property :}$

$\bf{\longmapsto \: \angle \: ADO + \angle \: BAD + \angle \: ABD = 180 ^{ \circ} }$

$\bf{ \longmapsto \: \angle \: ADO + {90}^{ \circ} + {31}^{ \circ} = {180}^{ \circ} }$

• We took ∠BAD = 90° as measure of all angles of rectangle is 90° and ∠ABD or ∠ABO = 31°.

$\bf{ \longmapsto \: \angle \: ADO + {121}^{ \circ} = {180}^{ \circ} }$

$\bf{ \longmapsto \: \angle \: ADO = {180}^{ \circ} - {121}^{ \circ} }$

$\bf{ \longmapsto \: \angle \: ADO = \pink{{59}^{ \circ}}}$

Hence,

➸ ∠ADO is 59°.

(iii) ∠OCB

➺ ∠OCB = ∠ADO [As we have already proved above]

➺ ∠ADO = 59° [As we already found it in (ii) part]

$\mathfrak{ \red{So,}}$

➸ ∠OCB = 59°

Hence,

➺ ∠OCB is 59°.