Question

ABCD is a trapezium with AB=AD, angle BCD= 50 degree and angle ABC=60 degree. Find a) angle ADB
b)angle CDB​

Answer 1

Step-by-step explanation:

if we exceed BA upto E then EAD =60°

then BAD=120

as AB=AD then in triangle ABD,

angle ABD=angle ADB÷x

x+x+120=180

2x=180-120=60

x=60/2=30

so angle ADB=30°

angle ABD=30°

angle DBC=60-30=30

angleBCD=50

I n triangle BDC,

angle DBC+angleBCD+angle CDB=180

30+50+angleCDB=180

angleCDB=180-80=100

Answer 2

Answer:

angle ADB = 30 degree angle CDB =100 degree

Step-by-step explanation:

Let angle BAD= angle 1 ; angle ADB = x ; angle BDC = y

AD = AB (given)

x = angle ABD = x (angles opposite to equal lines are equal)......... (1)

AD parallel to BC (a pair of opposite lines of a trapezium are parallel )

by co-interior angles

angle 1 + 60 = 180

angle 1 = 180 - 60

angle 1 = 120.......... (2)

in triangle ABD

by angle sum property of a triangle

angle 1 +x +x =180 (from 1)

120 +2x =180 (from 2)

2x =60

x = 60/2

x = 30

x +y +50 =180 (Co-interior angles )

30 + y +50=180

80 +y =180

y = 180-80

y = 100

Hope it will help you