Question

In the given figure angle OAB equal to 75 degree, angle OBA equal to 55 degree and angle OCD equal to 100 degree.Then angle ODC equal to?

Answer 1

Given :- In the given figure, ∠OAB is equal to 75°, ∠OBA is equal to 55° and ∠OCD is equal to 100°. Then ∠ODC is equal to ?

Solution :-

In ∆OAB, we have,

→ ∠OAB + ∠OBA + ∠AOB = 180° (By Angle sum Property.) → 75° + 55°+ ∠AOB = 180°

→ 130° + ∠AOB = 180°

→ ∠AOB = 180° - 130°

→ ∠AOB = 50°

now, as we can see that,

→ ∠COD = ∠AOB = 50° (vertically opposite Angles.)

In ∆OCD, we have,

→ ∠COD + ∠OCD + ∠ODC = 180° (By Angle sum Property.)

→ 50° + 100° + ∠ODC = 180°

→ 150° + ∠ODC = 180°

→ ∠ODC = 180° - 150°

→ ∠ODC = 30° (Ans.)

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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Answer 2

☆ Given :-

• ∠OAB = 75°
• ∠OBA = 55°
• ∠OCD = 100°

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☆ To find :-

• ∠ODC

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☆ Solution :-

$\large\underline\red{\bold{❥︎Step :- 1 }}$

☆ In Δ OAB, 

We know, ∠OAB + ∠OBA + ∠AOB = 180°    …(i)

[since, sum of all the angles of a triangle is 180°]

On putting ∠OAB = 75° and ∠OBA = 55° in eq. (i), we get

⇒ 75° + 55° + ∠AOB = 180°

⇒ 130° + ∠AOB = 180° 

⇒ ∠AOB = 180° -130° 

⇒ ∠AOB = 50°

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$\large\underline\purple{\bold{❥︎Step :- 2 }}$

Now, ∠AOB  = ∠COD [vertically opposite angles]

 ∴ ∠COD = 50°

In Δ COD, 

⇒ ∠COD + ∠OCD + ∠ODC = 180°   … (ii)

[since, sum of all the angles of a triangle is 180°]

On putting ∠OCD = 100° and ∠COD = 50° in eq. (ii), we get

⇒ 100° + 50° + ∠ODC = 180° 

⇒ 150° + ∠ODC = 180°

⇒  ∠ODC = 180° - 150°

 ∴  ∠ODC = 30°

Hence,∠ODC = 30°

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