23 Yuan Yeone a point p, two tangents PA cene. Po • P, ш. and PB are drawn to a cucce e coor). If OP=24 , Op then find (APB. What type of hangle is APB? 3
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InΔOAP:
OA=r
OP=2r
⇒PA
2
=OP
2
−OA
2
=4r
2
−r
2
=3r
2
⇒PA=
3
r
∴PA=PB=
3
r
(∵Tangents from an external point are equal in length )
Now, tan(∠APO)=
AP
OA
=
r
3
r
=
3
1
∠APO=30°
Similarly
∠BPO=30°
∴∠APB=60°
In ΔABP,
Using cosine rule :
cos(∠APB)=
2AP.BP
AP
2
+BP
2
−AB
2
⇒cos(60°)=
2(
3
r)
2
(
3
r)
2
+(
3
r)
2
−AB
2
⇒
2
1
=
2(3r
2
)
6r
2
−AB
2
⇒3r
2
=6r
2
−AB
2
⇒AB=
3
r
∴AB=AP=BP=
3
r
Hence, ABP is an equilateral triangle
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