question_answer

In a triangle PQR, PS is the angle bisector of \[\angle QPR\]and \[\angle QPS=60{}^\circ .\] What is the length of PS?

A)  \[\frac{\left( b+c \right)}{bc}\]               

B)  \[\frac{bc}{\left( b+c \right)}\]

C)  \[\sqrt{{{b}^{2}}+{{c}^{2}}}\]                    

D)  \[\frac{{{\left( b+c \right)}^{2}}}{bc}\]

Correct Answer: B

Solution :

Applying angle bisector theorm \[\frac{QS}{SR}=\frac{b}{c}\,\,\,\Rightarrow \,\,\,\,\frac{QS}{(a-QS)}=\frac{b}{c}\] \[QS=\frac{ab}{c}-QS\times \frac{b}{c}\Rightarrow \,\left( 1+\frac{b}{c} \right)\,QS=\frac{ab}{c}\] \[QS=\frac{ab}{b+c}\]                          ?..(i) In \[\Delta QPS,\,\,\,\frac{QS}{\sin 60{}^\circ }=\frac{PS}{\sin Q}\] (By using sine rule) \[PS=\frac{QS.\,\sin Q}{\sin 60{}^\circ }\]                                  ??(ii) In  \[\Delta PQR,\,\,\,\frac{a}{\sin 120{}^\circ }=\frac{c}{\sin Q}\] \[\sin Q=\frac{c}{a}\,\,\sin 120{}^\circ \] \[\therefore \,\,\,\,\,\,PS=\frac{\frac{ab}{b+c}\times \frac{c}{a}\times \sin 120{}^\circ }{\sin 60{}^\circ },\,\,\,\,PS=\underline{\frac{\mathbf{cb}}{\mathbf{b+c}}}\]