Question 810620
The sketch below shows the two buildings as green buildings rectangles.
{{{drawing(300,300,-60,440,-30,470,
line(0,150,307.55,150),
blue(line(0,150,307.55,0)),
blue(line(0,150,307.55,420.76)),
green(rectangle(0,150,-40,0)),
green(rectangle(307.55,0,400,420.76)),
line(-80,0,420,0),rectangle(0,0,15,15),
rectangle(307.55,150,292.55,135),rectangle(307.55,150,292.55,165),
red(arc(0,150,200,200,0,26)),red(arc(307.55,0,200,200,180,206)),
red(arc(0,150,150,150,-41.34,0)),locate(75,200,red(41.34^o)),
locate(100,150,red(26^o)),locate(170,50,red(26^o)),
locate(-10,175,A),locate(-15,25,X),locate(310,160,H),
locate(300,445,B),locate(310,25,Y)
)}}} {{{AX=150}}}{{{ft}}}
From the right triangles below horizontal line AH (triangles AHY and AXY) we know that
{{{150}}}{{{ft=AX=HY=XY*tan(26^o)}}} --> {{{AH=XY=150ft/tan(26^o)}}}
 
If we want to solve the problem in 3 steps,
with that we can calculate an approximate value for {{{AH=XY}}}= approximately {{{307.55ft}}} ,
and then we use that value and right triangle ABH to find the distance BH,
which we add to {{{HY=AX=150}}}{{{ft}}} to find {{{BY}}} , the height of the tall building.
 
If we want to be fancy, and do all of the calculations in one fancy formula,
we keep working with variables on right triangle ABH.
{{{BH=AH*tan(41.34^o)=150ft*tan(41.34^o/tan(26^o))=(150ft)*(tan(41.34^o)/tan(26^o))}}}
Then,
{{{BY=BH+HY=(150ft)*(tan(41.34^o)/tan(26^o))+150ft=highlight((150ft)*(1+tan(41.34^o)/tan(26^o)))}}}
That is the exact value answer that a mathematician would give.
 
A physics professor would tell you to consider the uncertainty in the measurements
(of the height of the shorter building, and the angles),
and may give an answer such as {{{421ft +- 2ft}}} .
 
A more practical engineer, calculates the approximate value as
{{{BY=420.757ft}}} or more likely {{{highlight(421ft)}}} .
I suspect that a high school math teacher would like {{{highlight(421ft)}}} as an answer too.