Sides BC, BA and CA of ∠ABC are produced to D, F and E respectively. If ∠ACD = 120° and ∠EAF = 45°, find three angles of ∠ABC.
Since lines CE and BF intersect at A.
⇒ ∠BAC = ∠EAF = 45° ….(Vertically opposite angles)
Now ∠ACD = ∠BAC + ∠ABC
⇒ 120° = 45° + ∠ABC
⇒ ∠ABC = 120° - 45° = 75°
Now in ∠ABC, we have
∠A + ∠B + ∠C = 180°
Þ 45° + 75° + ∠C = 180°
... ∠C = 180° - 45° - 75° = 60°
Hence three angles of ΔABC are 45°, 75° and 60°.
⇒ ∠BAC = ∠EAF = 45° ….(Vertically opposite angles)
Now ∠ACD = ∠BAC + ∠ABC
⇒ 120° = 45° + ∠ABC
⇒ ∠ABC = 120° - 45° = 75°
Now in ∠ABC, we have
∠A + ∠B + ∠C = 180°
Þ 45° + 75° + ∠C = 180°
... ∠C = 180° - 45° - 75° = 60°
Hence three angles of ΔABC are 45°, 75° and 60°.
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