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
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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
Answered by
11
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:
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