question_answer

The external bisector of ZB and ZC of AABC (where AB and AC extended to E and P respectively) meet at point P. If ZBAC =\[100{}^\circ \], then the measure of ZBPC is

A) \[50{}^\circ \]                          

B) \[80{}^\circ \]

C) \[40{}^\circ \]              

D)        \[100{}^\circ \]

Correct Answer: C

Solution :

 tan (2\[\theta \] + \[45{}^\circ \]) = cot 3\[\theta \] = tan \[\left( 90{}^\circ -3\theta  \right)\] \[\Rightarrow \]\[2\theta +45{}^\circ =90{}^\circ -3\theta \] \[\Rightarrow \]\[5\theta =90{}^\circ -45{}^\circ =45{}^\circ \]  \[\therefore \]\[\theta =9{}^\circ \]  AB = Length of the thred = 150 metre \[\angle \]BAC =\[60{}^\circ \] In \[\Delta \]ABC. sin \[60{}^\circ \] =\[\frac{BC}{AB}\Rightarrow \frac{\sqrt{3}}{2}=\frac{BC}{150}\] \[\Rightarrow \]BC \[=150\times \frac{\sqrt{3}}{2}=75\sqrt{3}\] metre