A) \[3:1\]
B) \[4:1\]
C) \[2:1\]
D) \[3:2\]
Correct Answer: C
Solution :
\[\angle BPA={{90}^{o}}\] (Angle in semicircle) In \[\Delta \,BPA,\,\angle ABP+\angle BPA+\angle PAB={{180}^{o}}\] \[\Rightarrow \] \[{{30}^{o}}+{{90}^{o}}+\angle PAB={{180}^{o}}\] \[\Rightarrow \] \[\angle PAB={{60}^{o}}\] Also. \[\angle POA=2\angle PBA\] \[\Rightarrow \] \[\angle POA=2\times {{30}^{o}}={{60}^{o}}\] (side opposite to equal angles) In \[\Delta \,\,OPT,\,\,\angle OPT={{90}^{o}}\] \[\angle POT={{60}^{o}}\]and \[\angle PTO={{30}^{o}}\] [angle sum property of a \[\Delta \]] Also \[\angle APT+\text{ }\angle ATP=\angle PAO\][exterior angle property] \[\therefore \] \[\angle APT+{{30}^{o}}={{60}^{o}}\Rightarrow \angle PT={{30}^{o}}\] \[\therefore \] AP=AT ...(is) (side opposite to equal angles) From (i) and (ii), AT = OP = radius of the circle; and AB = 2r \[\Rightarrow \] \[AB=2AT\Rightarrow \frac{AB}{AT}=2\Rightarrow AB:AT=2:1\]You need to login to perform this action.
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