Question 42272
{{{drawing(400,400,-25,375,-25,375,
line(0,0,350,0),
line(350,0,350,300),
line(0,0,350,300),
line(250,0,350,300),
locate(0,0,A),
locate(350,0,B),
locate(350,300,C),
locate(250,0,D)
)}}}


Let the points C and B denote the top and the base of the mountain respectively.
When the person was at D, the angle of elevation was {{{39^o}}} and when he moved to A, 100 m away from the mountain, the angle of elevation became {{{38.2^o}}}.


So, in the figure, AD = 100 m; < BAC = {{{38.2^o}}}; < BDC = {{{39^o}}}


In right-angled triangle ABC,
{{{BC/AB = tan(38.2^o)}}}
or {{{AB = BC/(tan(38.2^o))}}} _____(1)


In right-angled triangle DBC,
{{{BC/DB = tan(39^o)}}}
or {{{DB = BC/(tan(39^o))}}} _____(2)


Subtracting (2) from (1) we have
{{{AB - DB = BC/tan(38.2^o) - BC/tan(39^o)}}}
or {{{AD = BC(cot(38.2^o) - cot(39^o))}}}
or {{{BC = AD/(cot(38.2^o) - cot(39^o))}}}
or {{{BC = 100/(1.270773 - 1.234897)}}}
or BC = 2787.4 m (approx) = 2.7874 km


Hence, the reqd. height of the mountain is 2.7874 km.