In a rhombus ABCD, if ∠ACB =40°, then ∠ADB =
The diagonals in a rhombus are perpendicular,
So,
∠BPC = 90o
From triangle BPC,
The sum of angles is 180°
So,
∠CBP = 180o – 40o – 90o
= 50°
Since, triangle ABC is isosceles
We have,
AB = BC
So,
∠ACB = ∠CAB = 40o
Again from triangle APB,
∠PBA = 180o – 40o – 90o
= 50o
Again, triangle ADB is isosceles,
So,
∠ADB = ∠DBA = 50o
∠ADB = 50o