in the figure find the values of X and Y
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∠CBD is exterior angle of ∆ABC
∠CBD=∠CAB+∠ACB
∠y=50°+30°
∠y=80°
Let point P be at ∠x
In ∆PBD,
∠B+∠D+∠P=180°
∠y+55+∠P=180
80+55+∠P=180
135+∠P=180
∠P=180-135=45°
∠P+∠x=180(linear pair)
45+∠x=180
∠x=180-45=135
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