Question

In a figure given below ,ABCD is a parallelogram in which angle DAB=70,angle =80. calulate angle CDB and ADB

Answer 1

which angle is 80 in your question

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