If the angles of elevation of the top of a tower from three collinear points A,B and C, on a line leading to the foot of the tower, are \[{30^ \circ }\], \[{45^ \circ }\] and \[{60^ \circ }\] respectively, then the ratio, AB:BC, is :
Solutions · 1
Official solution
As PB bisects \[\angle APC,\] therefore AB:BC=PA:PC
Also in \[\Delta APQ,\sin {30^ \circ } = {h \over {PA}} \Rightarrow PA = 2h\]
and in \[\Delta CPQ,\] \[\sin {60^ \circ } = {h \over {PC}} \Rightarrow PC = {{2h} \over {\sqrt 3 }}\]
\[\therefore\] \[AB:BC = 2h:{{2h} \over {\sqrt 3 }} = \sqrt 3 :1\]