Question

In the adjacent figure, it is given that AB =AC, ∠BAC=36°,∠ADB=45° and ∠AEC = 40°. Find (i) ∠ABC (ii) ∠ACB (iii) ∠DAB (iv) ∠EAC.

Answer 1

(I). 72°
(ii). 72°
(iii). 27°
(iv). 32°

Answer 2

Answer:

∠ACB=72° , ∠ABC =72° , ∠DAB=27° and ∠EAC=32°

Step-by-step explanation:

In ΔABC

AB = AC

Since we know that the opposite angles of equal sides are equal

So, ∠ACB=∠ABC  ---1

Now By angle sum property of triangle :

∠ABC+∠ACB +∠BAC = 180°

using 1

∠ABC+∠ABC +∠BAC = 180°

2∠ABC +36° = 180°

2∠ABC = 180°-36°

2∠ABC = 144°

∠ABC = 72°

So, ∠ACB=∠ABC =72°

Now  ∠BAC+∠ACB=∠ABD (exterior angle property of triangle )

36°+72° =∠ABD

108° =∠ABD

In ΔABD

∠ADB+∠ABD+∠DAB=180° (Angle sum property of triangle )

45°+108°+∠DAB=180°

153°+∠DAB=180°

∠DAB=180° -153°

∠DAB=27°

Now  ∠BAC+∠ABC=∠ACE (exterior angle property of triangle )

36°+72° =∠ACE

108° =∠ACE

In ΔACE

∠AEC+∠ACE+∠EAC=180° (Angle sum property of triangle )

40°+108°+∠DABEAC=180°

148°+∠EAC=180°

∠EAC=180° -148°

∠EAC=32°

Hence ∠ACB=72° , ∠ABC =72° , ∠DAB=27° and ∠EAC=32°