Question

Please solve this urgently

Answer 1

Answer:

$Construction:~Join~A,C~and~P,Q.~AC~and~BE~intersect~at~G$

                         $and~PQ~and~BE~intersect~at~H$

$BG$⊥$AC~and~AG=GC [BG~is~bisector~of~$

$In~right~angled~triangle~ABG,\\sinABG = \frac{AG}{AB}\\or, sin15^o = \frac{AG}{x}\\or, AG = xsin15^o=GC\\Also,~PH=HQ=xsin15^o [APHG ~and~ GHQC~ are~congruent~rectangles.]\\In~right~angled~triangle~PBH,\\ sinPBH = \frac{PH}{PB}\\ or, sinPBH = \frac{xsin15^o}{10}$$Now,\\cosPBQ = cos[2(PBH)]~~~~~[$