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प्रश्न
In triangle ABC, D is a point in AB such that AC = CD = DB. If ∠B = 28°, find the angle ACD.
उत्तर
ΔDBC is an isosceles triangle.
As, Side CD = Side DB
⇒ ∠DBC = ∠DCB ...[If two sides of a triangle are equal, then angles opposite to them are equal]
And ∠B = ∠DBC = ∠DCB = 28°
As the sum of all the angles of the triangle is 180°
∠DCB + ∠DBC + ∠BCD = 180°
⇒ 28° + 28° + ∠BCD = 180°
⇒ ∠BCD = 180° − 56°
⇒ ∠BCD = 124°
Sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.
⇒ ∠DBC+ ∠DCB = ∠ADC
⇒ 28° + 28° = ∠ADC
⇒ ∠ADC = 56°
Now ΔACD is an isosceles triangle with AC = DC
⇒ ∠ADC = ∠DAC = 56°
Sum of all the angles of a triangle is 180°
⇒ ∠ACD + ∠ADC + ∠DAC = 180°
⇒ ∠ACD + 56° + 56° = 180°
⇒ ∠ACD = 180° − 112°
⇒ ∠ACD = 68°
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