Question

In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are

Answer 1

Step-by-step explanation:

Answer :

Given,

ABCD is a rhombus

∠ACB = 40°

∵ ∠DAC =∠ACB = 40° [alternate angles]

∠AOD = 90° [ diagonals of rhombus are perpendicular to each other]

In ∆AOD

∠AOD+∠ADO+∠DAO = 180°

∠ADO+90°+40° = 180°

= ∠ADO = 180°- 130° = 50°

=∠ADO=∠ADB = 50°

Hope it helps

Answer 2

Answer:

Step-by-step explanation:

Answer :

Given,

ABCD is a rhombus

∠ACB = 40°

∵ ∠DAC =∠ACB = 40° [alternate angles]

∠AOD = 90° [ diagonals of rhombus are perpendicular to each other]

In ∆AOD

∠AOD+∠ADO+∠DAO = 180°

∠ADO+90°+40° = 180°

= ∠ADO = 180°- 130° = 50°

=∠ADO=∠ADB = 50°

Step-by-step explanation:

Hopes it's helpful

Mark as brainlist ...

keep smiling dear...