Question

In fig 9,ABCD is a rhombus in which angle ABD=40.find angle BAC angle BCD angle ADC

Answer 1

Diagonal of rhombus bisect each othe at 90 degree
so in triangle AOB
we get
AOB+OAB+ABO=180
90+OAB+40=180
AngleOAB=AngleBAC=50

angleBDC = angleABD
BDC=40

angleBAC=angleACD
ACD=50

Answer 2

Answer:

Angles of a Rhombus

Answer: On Summarizing our findings we get

1. ∠ CAB = 50° and

2.∠BCD = 100°

3.∠ADC = 80°

Explanation:

Given that ABCD is a RHOMBUS

and ∠ABD = 40°

Need to determine ∠CAB , ∠BCD and ∠ADC

One of the important property of RHOMBUS IS DIAGONAL OF THE RHOMBUS BISECTS OPPOSITE ANGLES.

⇒ ∠ABD = ∠DBC = (1/2)∠ABC = 40° and

also ∠ CAB = (1/2)∠DAB   ---------eq(1)

As (1/2)∠ABC = 40°

⇒ ∠ABC = 40° × 2 = 80°

since OPPOSITE ANGLES OF RHOMBUS ARE EQUAL

⇒ ∠ADC = ∠ABC = 80°

⇒ ∠ADC = 80°

And also CONSECUTIVE ANGLES OF RHOMBUS WHICH IS A KIND OF PARALLEOGRAM ARE SUPPLEMETARY

⇒ ∠ADC  + ∠BCD = 180°

⇒ 80° + ∠BCD = 180°       [ Since ∠ADC = 80° ]

⇒ ∠BCD = 180° - 80° = 100°

⇒ ∠BCD = 100°

⇒ ∠DAB  = ∠BCD = 100°     [ opposite angles of Rhombus are equal ]

⇒∠DAB  =  100°

from eq (1) ∠ CAB = (1/2)∠DAB   = (1/2)×80° = 50°

⇒ ∠ CAB = 50°

On Summarizing our findings we get

1. ∠ CAB = 50° and

2.∠BCD = 100°

3.∠ADC = 80°