question_answer

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. lf \[\angle PBT={{30}^{o}},\] then \[BA:AT\]is                        

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 \[\Rightarrow \] \[AB=2AT\Rightarrow \frac{AB}{AT}=2\Rightarrow AB:AT=2:1\]