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Question

Triangle ABC has AB=90,BC=50 and CA=70. A circle is drawn with centre P on AB such that CA and CB are tangents to the circle. Find 23AP. 
(correct answer + 3, wrong answer 0)


Solution

Let CA touches the circle at R and CB touches the circle at S.
Let Q be a point on AB so that CQ and AB are perpendicular.


Let r be the radius of the circle.
From similar triangles AQC and ARP,
CQr=70AP     (1)
From similar triangles BQC and BSP,
CQr=50BP=50(90AP)     (2)

From (1) and (2),
7(90AP)=5AP
630=12AP
2AP=105
Therefore, 23AP=1053=35.

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