Question

11. In the adjoining figure, ABCD is a parallelogram in which angle BAD = 75° and angle DBC = 60°. Calculate: (i) angle CDB and (ii) angle ADB. ​

Answer 1

Answer:

Here's your answer :-

Step-by-step explanation:

In ||gm ABCD,

we know that AB || CD

● angle A = anlge C ( in ||gm opposite angles are equal)

=> angle C = 75°

Let angle CDB = x°

By angle sum property in triangle CDB

C + CDB + CBD = 180°

75 + x + 60 = 180

135 + x = 180

x = 180 - 135

x = 45°

So, angle CDB = x = 45°

Now, as AB || CD and taking BD as trasversal.

angle CDB = angle ABD (alternate interior angles)

=> angle ABD = 45°

By angle sum property in triangle ABD

A + ABD + ADB = 180°

75 + 45 + ADB = 180

120 + ADB = 180

ADB = 180 -120

ADB = 60°

Answer 2

Answer:

We know that the opposite angles of a parallelogram are equal , so angle BCD = BAD= 75 degree

(i) Now,in triangle BCD,we have

angle CDB +DBC+BCD=180 degree( because sum of the angles of a triangle is 180 degree)

angle CDB+60 degree+75 degree =180 degree

angle CDB+135 degree=180 degree

angle CDB=( 180 degree-135 degree)=45 degree

(ii) AD||BC and BDis a transversel

because angle ADB =angle DBA =60 degree

Hence, angle ADB = 60 degree