Question

In the given figure, angleCAB = 40⁰,AC=AB and BC = BD. Find:
¡] angleACB
¡¡] angleCDB ​

Answer 1

Answer:

i) since sides AC=BC

therefore it is an isosceles triangle

angle ABC = angle ACB ( eqation 1)

now angle ABC + BCA + CAB = 180' (angle sum)

40° + ACB + ACB = 180 ( from eqation 1)

2ACB = 180 - 40

ACB = 140/ 2

ACB = 70°

Answer 2

Step-by-step explanation:

In triangleBAC

now,

AB=AC (given)

so also, angleABC=angleACB = a

then, 40°+a+a=180°

40+2a=180°

2a=140°

a=140/2

a=70°

so,angleACB=70°

Now,

angleABC+angleCBA=180° (linear pair)

70°+angleCBA=180°

angleCBA=110°

SO, Side CB=BD

Also, angleBCD=angleBDC= z

Now 110+2z= 180°

2z=70°

z=35°

so angleCDB=35°

I HOPE THIS HELP....

MARK AS BRILIENT