in the fig,<ACD is an exterior angle of∆ABC,<B=50°,<A=60°,find the measure of <ACD
Answers
Answer:
In figure 3.8, ∠ACD is an exterior angle of ΔABC. ∠B =40°,∠A=70°.
the measure of ∠ACD = 110°
in a triangle sum of all angles = 180°
∠ABC + ∠BAC + ∠ACB = 180°
=> ∠B + ∠A + ∠ACB = 180°
=> 40° + 70° + ∠ACB = 180°
=> ∠ACB = 70°
∠ACB + ∠ ACD = 180° ( straight line)
=> 70° + ∠ ACD = 180°
=> ∠ ACD = 110°
Step-by-step explanation:
Given : side BC of ∆ ABC is extended up to the point D so that ∠ACD is exterior angle
∠ A = 50° and ∠ B = 60°
To Find : the measure of ∠ACD is
Solution:
Sum of angles of a triangle is 180°
=> ∠ A + ∠ B + ∠ C = 180°
Substitute ∠ A = 50° and ∠ B = 60°
=> 50° + 60° + ∠ C = 180°
=> ∠ C + 110° = 180°
∠ C is ∠ACB
=> ∠ACB + 110° = 180°
Now ∠ACB and ∠ACD form a linear pair as BC is extended to D
=> ∠ACB + ∠ACD = 180°
Equate both Equations:
∠ACB + ∠ACD = ∠ACB + 110°
=> ∠ACD = 110°
Hence the measure of ∠ACD is 110°
Shortcut:
Exterior angle of Triangle = Sum of opposite two interior angles
Hence ∠ACD = 50° + 60° = 110°
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