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 \[\Delta PQR,\,\,\,\frac{a}{\sin 120{}^\circ }=\frac{c}{\sin Q}\] \[\sin Q=\frac{c}{a}\,\,\sin 120{}^\circ \] \[\therefore \,\,\,\,\,\,PS=\frac{\frac{ab}{b+c}\times \frac{c}{a}\times \sin 120{}^\circ }{\sin 60{}^\circ },\,\,\,\,PS=\underline{\frac{\mathbf{cb}}{\mathbf{b+c}}}\]You need to login to perform this action.
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