Question

In figure, the sides BC, CA and BA of a ΔABC have been produced to D, E and F respectively. If ∠ACD=105°
and ∠EAF=45°
; find all the angles of the ΔABC.​

Answer 1

Given: ∠ACD=105°

,∠EAF=45°

To find: All angles of △ABC

Solution: Clearly, ∠BAC=∠EAF=45°

[Vertically opposite angle]

and, ∠ACB+∠ACD=180°

[Linear pair]

⇒∠ACB+105°

=180°

⇒∠ACB=180°

−105°

⇒∠ACB=75°

Now, in △ABC by angle sum property of triangle, we have

∠BAC+∠ABC+∠ACB=180°

⇒45°

+∠ABC+75∘

=180°

⇒∠ABC=180∘

−45∘

−75°

⇒∠ABC=60°

Answer 2

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In a △ABC,

∠BAC and ∠EAF are vertically opposite angles.

Hence,

we can write as

∠BAC = ∠EAF = 45°

Considering the exterior angle property,

we have

∠BAC + ∠ABC = ∠ACD = 105°

On rearranging we get

∠ABC = 105°– 45° = 60°

We know that the sum of angles in a triangle is 180°

∠ABC + ∠ACS +∠BAC = 180°

∠ACB = 75°

Therefore, the angles are 45°, 65° and 75°.