Question 894562
Here's the figure:
{{{drawing(100,350,-80,20,-15,335,
line(-80,0,20,0),
rectangle(0,0,10,327.4),
red(line(-59.9,0,0,327.4)),
green(line(-35.9,0,0,327.4)),
rectangle(-8,0,0,8),
locate(-55,14,24),locate(-21,14,x),
locate(-8,170,y)
)}}} It has two right triangles: {{{drawing(100,350,-80,20,-15,335,
line(-80,0,20,0),
rectangle(0,0,10,327.4),
red(line(-59.9,0,0,327.4)),
red(arc(-59.9,0,30,30,-79.63,0)),
locate(-52,40,red(79.63^o)),
rectangle(-8,0,0,8),
locate(-45,0,x+24),
locate(-8,170,y)
)}}} and {{{drawing(100,350,-80,20,-15,335,
line(-80,0,20,0),
rectangle(0,0,10,327.4),
green(line(-35.9,0,0,327.4)),
green(arc(-39.9,0,30,30,-83.74,0)),
locate(-52,40,green(83.74^o)),
rectangle(-8,0,0,8),
locate(-21,0,x),
locate(-8,170,y))}}}
You can re-write the equations as
{{{y=(x+24)*tan(79.63^o)=x*tan(83.74^o)}}} , which is really 3 equations in one.
 
{{{(x+24)*tan(79.63^o)=x*tan(83.74^o)}}}-->{{{x*tan(79.63^o)+24*tan(79.63^o)=x*tan(83.74^o)}}}-->{{{24*tan(79.63^o)=x*tan(83.74^o)-x*tan(79.63^o)}}}-->{{{24*tan(79.63^o)=x*(tan(83.74^o)-tan(79.63^o))}}}-->{{{24*tan(79.63^o)/(tan(83.74^o)-tan(79.63^o))=x}}}
 
At this point, using the approximate values {{{system(tan(79.63^o)=5.464685,tan(83.74^o)=9.116232)}}} , you could calculate {{{x=35.917}}}(rounded),
but you do not need to report/write the value for {{{x}}} , so you can keep calculating (just multiply times {{{tan(83.74^o)=9.116232}}} ) to get {{{y}}} as shown below.
 
{{{system(y=x*tan(83.74^o),x=24*tan(79.63^o)/(tan(83.74^o)-tan(79.63^o)))}}} ---> {{{y=24*tan(79.63^o)*tan(83.74^o)/(tan(83.74^o)-tan(79.63^o))}}}
Using the approximate values {{{system(tan(79.63^o)=5.464685,tan(83.74^o)=9.116232)}}} ,
{{{y=24*5.464685*9.116232/(9.116232-5.464685)}}} ---> {{{highlight(y=327.4)}}} (rounded).
The height of the building is 327 meters.