Question

in the adjoining figure ABCD is a parallelogram and diagonals intersect at O find angle C A D,angle ACD and angle ADC.​

Answer 1

a)∠CAD = 66°

b)∠ACD = 38°

c)∠ADC= 76°

Step-by-step explanation:

Given,ABCD is a parallelogram.

By considiring the attached figure,

In ΔBCD,

∠CDB + ∠CBD + ∠BCD = 180° [Angle sum property of triangle]

∠BDC = 30°, ∠CBD = 46°

So,30° + 46° +∠BCD = 180°

⇒76° + ∠BCD = 180°

⇒∠BCD = 180 - 76

              = 104°

here,ACD is a straight line

⇒∠AOD + ∠COD = 180° [By linear pair property]

⇒68°+∠COD =180°

⇒∠COD = 180°-68°

              = 112°

Now,In ΔCOD,

∠ODC +∠OCD +∠COD =180° [Angle sum property of a triangle]

30°+∠OCD + 112° =180°

142° + ∠OCD =180°

∠OCD = 180°-142°

          =38°

⇒∠ACD = 38°

In ΔBOC,

∠CBO+∠BCO+∠BOC =180° [Angle sum property of a triangle]

∠AOD = ∠BOC =68°(Vertically opposite Angles)

46°+∠BCO+68° =180°

114°+∠BCO = 180°

∠BCO = 180°-114°

          = 66°

Here ∠BCO = ∠CAD [Alernate interior Angles]

⇒∠CAD = 66°

Now,In ΔΔADC,

∠ADC+∠CAD+∠ACD = 180° [Angle sum property of a triangle]

∠ADC +  66°+ 38°= 180°

∠ADC+ 104° = 180°

∠ADC = 180°-104°

           = 76°

a)∠CAD = 66°

b)∠ACD = 38°

c)∠ADC= 76°

*Refer the attached figure.

Answer 2

Answer:

hey mate please check the pictures that are given you will find your answer there and if any problem then ask me in comment section.

Thank you.