Q. 24.0( 8 Votes )
In the figure, if ∠BAC = 80° then find the value of ∠BCP.
Answer :
AC is the diameter
∠ ACP = 90° [radius is perpendicular to tangent]
∠ ABC = 90° [angle in a semicircle is 90°]
∠ ABC + ∠ BAC + ∠ ACB = 180° [sum of internal angles in a triangle is 180°]
∠ ACB + 90° + 80° = 180°
∠ ACB = 180 – 170
∠ ACB = 10°
∠ BCP = ∠ ACP - ∠ ACB
= 90 – 10
= 80°
∴ ∠ BCP = 80°
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RELATED QUESTIONS :
In the figure, PQ and XY are parallel tangents. If ∠QRT = 30°, then find the value of ∠TSY.
In the figure, if ∠BAC = 80° then find the value of ∠BCP.
