Question

in the given figure ABCD is a parallelogram in which angle DAB =80 and angle DBC=60 , Compute angle CDB AND angle ADB​

Answer 1

Step-by-step explanation:

ABCD is a parallelogram

Angle DAB = 80

Angle DBC = 60

To find angle CDB and angle ADB

Angle ADB = Angle DBC (They are alternate angles

Angle ADB = 60

Now,

Angle A + Angle D = 180

80 + 60 + angle CDB = 180

angle CDB = 180 - 140

angle CDB = 40

Answer 2

(a) Since, ABCD is a || gm

We have, AB || CD

∠ADB = ∠DBC (Alternate angles)

∠ADB = 80o (Given, ∠DBC = 80o)

Now,

In ∆ADB, we have

∠A + ∠ADB + ∠ABD = 180o (Angle sum property of a triangle)

70o + 80o + ∠ABD = 180o

150o + ∠ABD = 180o

∠ABD = 180o – 150o = 30o

Now, ∠CDB = ∠ABD (Since, AB || CD and alternate angles)

So,

∠CDB = 30o

Hence, ∠ADB = 80o and ∠CDB = 30o.

(b) Given, ∠BOC = 35o and ∠CBO = 77o

In ∆BOC, we have

∠BOC + ∠BCO + ∠CBO = 180o (Angle sum property of a triangle)

∠BOC = 180o – 112o = 68o

Now, in || gm ABCD

We have,

∠AOD = ∠BOC (Vertically opposite angles)

Hence, ∠AOD = 68o.

(c) ABCD is a rhombus

So, ∠A + ∠B = 180o (Sum of adjacent angles of a rhombus is 180o)

72o + ∠B = 180o (Given, ∠A = 72o)

∠B = 180o – 72o = 108o

Hence,

x = ½ B = ½ x 108o = 54o