Question

in triangle ABC ,AB divides angle DAC in the ratio of 1:3and AB=DB,DA extend to E Such that angle EAC=180°,angle ABC=36° then angle ACB is​

Answer 1

Answer:

ANSWER

∠DAC=180°−108°=72°∠DAB∠BAC=31 Given

⇒∠DAB=3∠BAC∴∠BAC+∠DAB=∠DAC=72°⇒∠BAC+3∠BAC=72°⇒4∠BAC=72°⇒∠BAC=472=18°∴∠DAB=3×18°=54°∴∠DAB=∠BDA=54°(AB=DB)∴∠ABD=180°−(54°+54°)=180−108°=72°

No, ∠DBA=72°=∠BAC+x (Exterio angle)

⇒x=72°−18°=54°

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