Answer:
[tex]Construction:~Join~A,C~and~P,Q.~AC~and~BE~intersect~at~G[/tex]
[tex]and~PQ~and~BE~intersect~at~H[/tex]
[tex]BG[/tex]⊥[tex]AC~and~AG=GC [BG~is~bisector~of~<ABC][/tex]
[tex]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} \\[/tex][tex]Now,\\cosPBQ = cos[2(PBH)]~~~~~[<PBH = <QBH]\\~~~~~~~~~~~~=1-2sin^{2}PBH\\~~~~~~~~~~~~=1-2(\frac{xsin15^o}{10})^2\\~~~~~~~~~~~~= 1-\frac{x^{2}sin^{2}15}{50}\\~~~~~~~~~~~~=1-\frac{x^2(2-\sqrt{3}) }{4(50)} \\~~~~~~~~~~~~=1-\frac{2-\sqrt{3} }{200} x^2[/tex]
