Question 42271
{{{drawing(200,200,-10,45,-5,30,
line(0,0,39,0),
line(39,0,39,25),
line(39,9,0,0),
line(0,0,39,25),
locate(0,0,A),
locate(39,0,B),
locate(39,9,C),
locate(39,25,D)
)}}}

Let CD be the tower on the inclined plane AC.
Inclination of AC with ground is {{{13^o}}} is < BAC = {{{13^o}}}.
Angle of elevation of the top of the tower is {{{20^o}}} from the point A which is at a distance of 40 m along the inclined plane from the base C of the tower.
So < CAD = {{{20^o}}} and AC = 40 m.
So < BAD = < BAC + < CAD = {{{13^o + 20^o}}} = {{{33^o}}}


In right-angled triangle ABC,
{{{AB/AC = cos(13^o)}}}
or {{{AB = AC*cos(13^o)}}}
or {{{AB = 40*cos(13^o)}}} _____(1)


In right-angled triangle ABD,
{{{BD/AB = tan(33^o)}}}
or {{{BD = AB*tan(33^o)}}}
or {{{BD = 40*cos(13^o)*tan(33^o)}}}  [from (1)]


Also, in right-angled triangle ABD, 
{{{BC/AC = sin(13^o)}}}
or {{{BC = AC*sin(13^o)}}}
or {{{BC = 40*sin(13^o)}}}


Height of the tower = CD = BD - BC 
= {{{40*cos(13^o)*tan(33^o) - 40*sin(13^o)}}}
= 16.31


Hence, the reqd. height of the mountain is 16.31 m.