In given figure, ABCD is a rhombus in which angle ABD = 40^o.

Find:
i. $∠BAC$
ii. $∠BDC$
iii. $∠ACD$"

Given: ABCD is a rhombus in which angle ABD = 40o.

To find: Here we have to find the value the value ∠BAC, ∠BCD and ∠ADC.

Solution:

Diagonal of a rhombus bisect each other at 90^o:

So, in ∆AOB:

∠BOA $+$ ∠OAB $+$ ∠ABO = 180^o  

90^o $+$ ∠OAB $+$ 40^o = 180^o 

∠OAB = 180^o $-$ (90^o $+$ 40^o)

∠OAB = 50^o 

i)

∠BAC = ∠OAB

∠BAC = 50^o 

ii)

∠BDC = ∠ABD   (Alternate interior angles)

∠BDC = 40^o

iii)

∠ACD = ∠BAC   (Alternate interior angles) 

∠ACD = 50^o 

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Updated on: 10-Oct-2022

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