Question

ABCD is a rhombus such that ∠ACB = 40°, then ∠ADB is

Answer 1

${ \underline{ \boxed{\sf{\Large{\red{\implies{ \: 50°}}}}}}}$

${\sf{\Large{\gray{{ \: Explanation:-}}}}}$

${\sf{\pink{{ \: Given :}}}}$

• ABCD is a rhombus.
• ∠ACB = 40°

${\sf{\pink{{ \: To \: find :}}}}$

• ∠ADB

${\sf{\pink{{ \: Solution :}}}}$

ABCD is a rhombus,

∠ACB = 40°

∵ ∠ACB = 40°

⇒ ∠OCB = 40°

AD || BC then,

⇒ ∠DAC = ∠BCA = 40° [Alternate interior angles]

⇒ ∠DAO = 40°

[ Since, diagonals of a rhombus are perpendicular to each other.]

We have,

∠AOD = 90°

We know that, Sum of all angles of a triangle is 180°:-

⇒ ∠AOD + ∠ADO + ∠DAO = 180°

⇒ 90° + ∠ADO + 40° = 180°

⇒ 130° + ∠ADO = 180°

⇒ ∠ADO = 180° – 130°

⇒ ∠ADO = 50°

⇒ ∠ADB = 50°

Hence, ∠ADB = 50°