Given, AB=10cm, Let OB=OA=r,OP=2OA=2r △OAP is right angle at A. OA2+AP2=OP2 [∵( Base )2+( Perpendicular )2 =( Hypotenuse )2] (AP)2=(OP)2−(OA)2 =(2r)2−(r)2=4r2−r2 ⇒(AP)2=3r2 AP=r√3 . . . (i) In △OAP,tanθ=
OA
AP
=
r
r√3
⇒tanθ=
1
√3
=tan30∘ θ=30∘ ∠OAP+∠APO+∠POA=180∘ 90∘+30∘+∠POA=180∘ ∠POA or α=180∘−120∘=60∘ sinα=