Q2 of 12 Page 167

In the figure, if ∠BAC = 80° then find the value of ∠BCP.

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°


More from this chapter

All 12 →
8

In the figure, O is the centre of the circle. Tangents PA and PB drawn from an external point P touch the circle at A and B respectively. Prove that PAOB is a cyclic quadrilateral.

 1

Answer the following questions from the figure.


(i) Alternate segment of ∠BAQ.


(ii) Alternate segment of ∠DAP.


(iii) If C is jointed to B, then ∠ACB is equal to which angles?


(iv) ∠ABD and ∠ADB are equal to which angles?

 3

In the figure, PQ and XY are parallel tangents. If ∠QRT = 30°, then find the value of ∠TSY.

 4

In the figure, the diagonal AC or cyclic quadrilateral ABCD bisects ∠C. Prove that the diagonal BD is parallel to the tangent at point A of the circle passing through A, B, C and D.