Selina solutions

Selina solutions

Grade 6

Angles (With their Types) | Exercise 24(A)

Question 13

Q13)

In the given figure :

(i) If ∠AOB = 45°, ∠BOC = 30° and ∠AOD= 110°;

find : angles COD and BOD.

(ii) If ∠BOC = ∠DOC = 34° and ∠AOD = 120° ;

find : angle AOB and angle AOC.

(iii) If ∠AOB = ∠BOC = ∠COD = 38°

find : reflex angle AOC and reflex angle AOD.

Solution 13:

(i) \angle COD=\angle AOD-\angle AOC

=\angle AOD-\left(\angle AOB+\angle BOC\right)

=110° -(45° +30°)

=110°-75°=35°

\angle BOD=\angle AOD-\angle AOB

=110°-45°

=65°

(ii) \angle AOB=\angle AOD-\angle BOD

=\angle AOD-\left(\angle BOC+\angle COD\right)

=120° - (34°+34°)

=120° -68°

=86°

(iii) Reflex \angle AOC=360° - \angle AOC

=360° - \left(\angle BOC+\angle COD\right)

=360° -(38° +38°)

=360° -76° =284°

Reflex \angle AOD=360° -\angle AOD

=360° \left(\angle AOB+\angle BOC+\angle COD\right)

=360° -(38° +38° +38° )

=360° -114°

=246°

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Question 13

Q13)

In the given figure :

(i) If ∠AOB = 45°, ∠BOC = 30° and ∠AOD= 110°;

find : angles COD and BOD.

(ii) If ∠BOC = ∠DOC = 34° and ∠AOD = 120° ;

find : angle AOB and angle AOC.

(iii) If ∠AOB = ∠BOC = ∠COD = 38°

find : reflex angle AOC and reflex angle AOD.

Solution 13:

(i) \angle COD=\angle AOD-\angle AOC

=\angle AOD-\left(\angle AOB+\angle BOC\right)

=110° -(45° +30°)

=110°-75°=35°

\angle BOD=\angle AOD-\angle AOB

=110°-45°

=65°

(ii) \angle AOB=\angle AOD-\angle BOD

=\angle AOD-\left(\angle BOC+\angle COD\right)

=120° - (34°+34°)

=120° -68°

=86°

(iii) Reflex \angle AOC=360° - \angle AOC

=360° - \left(\angle BOC+\angle COD\right)

=360° -(38° +38°)

=360° -76° =284°

Reflex \angle AOD=360° -\angle AOD

=360° \left(\angle AOB+\angle BOC+\angle COD\right)

=360° -(38° +38° +38° )

=360° -114°

=246°

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