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

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Solution

$In ∆ ABC, ∠ BAC and ∠ EAF are vertically opposite angles . Hence, we can say that : ∠ BAC = ∠ EAF = 45 ° Considering the exterior angle property, we can say that : ∠ BAC + ∠ ABC = ∠ ACD = 105 ° \Rightarrow ∠ ABC = 105 ° - 45 ° = 60 ° Because of the angle sum property of the triangle, we can say that : ∠ ABC + ∠ ACB + ∠ BAC = 180 ° ∠ ACB = 75 ° Therefore, the angles are 45 °, 60 ° and 75 ° .$

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