Question 1205123
<pre>
The solution above is incorrect. 

Let the tower be CT. It's a very steep hill.
The hill is AC. The observer is down at A.

{{{drawing(2000/7,400,-10,90,-10,130,

locate(25,40,8^o), locate(39,45,120),locate(81,104,30),
locate(75,110,theta),locate(0,0,A), locate(81.1,0,B),locate(79,125,T),
locate(81.1,92,C),

line(0,0,80.02660912,0), line(80.02660912,0,80.02660912,119.4189121),
line(0,0,80.02660912,89.41891206),line(0,0,80.02660912,119.4189121))}}}

We use the law of sines on triangle ACT

{{{120^""/sin(theta^"")}}}{{{""=""}}}{{{30^""/sin(8^o)}}} 

{{{30sin(theta^"")}}}{{{""=""}}}{{{120sin(8^o)}}}

{{{sin(theta^"")}}}{{{""=""}}}{{{120sin(8^o)/30^""}}}

{{{sin(theta^"")}}}{{{""=""}}}{{{4sin(8^o)}}}

{{{sin(theta^"")}}}{{{""=""}}}{{{0.5566924038}}}

This is the ambiguous case ASS, but since we know &theta; is acute,

{{{theta}}}{{{""=""}}}{{{33.82736264^o}}}

Angle &theta; and angle BAT are complementary, so

Angle BAT = {{{90^o}}}{{{""-""}}}{{{33.82736264^o}}}{{{""=""}}}{{{56.17263736^o}}}

So the angle of inclination of the hill is angle BAC

Angle BAC = Angle BAT - 8<sup>o</sup> = 56.17263736<sup>o</sup> - 8<sup>o</sup> = 48.17263736<sup>o</sup>.

Edwin</pre>