using the sine and cosine rules

Feb 2009
63
0
a tower 42 metres high, stands on top of a hill. From a point some distance from the base of the hill, the angle of elevation to the top of the tower is 13.2 degress. From the same point the angle of elevation to the bottom of the tower is 8.3 degrees. Find the height of the hill.
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
a tower 42 metres high, stands on top of a hill. From a point some distance from the base of the hill, the angle of elevation to the top of the tower is 13.2 degress. From the same point the angle of elevation to the bottom of the tower is 8.3 degrees. Find the height of the hill.
let \(\displaystyle h\) = height of the hill

\(\displaystyle x\) = horizontal distance from the observation point to a point directly beneath the tower

\(\displaystyle \tan(13.2) = \frac{h+42}{x}\)

\(\displaystyle \tan(8.3) = \frac{h}{x}\)

dividing the first equation by the second ...

\(\displaystyle \frac{\tan(13.2)}{\tan(8.3)} = \frac{h+42}{h}\)

solve for \(\displaystyle h\)
 
Nov 2009
717
133
Wahiawa, Hawaii
a tower 42 metres high, stands on top of a hill. From a point some distance from the base of the hill, the angle of elevation to the top of the tower is 13.2 degress. From the same point the angle of elevation to the bottom of the tower is 8.3 degrees. Find the height of the hill.
you can also solve this using the law of sines.

the triangle representing the top sight line, lower sight line, height of tower, one can drive all 3 interior angles \(\displaystyle 4.9^o,98.3^o,17.80^o\)

then the length of top sight line \(\displaystyle (\delta)\)is

\(\displaystyle
\frac{\sin{98.3^o}}{\delta} = \frac{\sin{4.9^o}}{42}
\rightarrow \delta=486.62
\)

now the hieght of the tower + hill becomes
\(\displaystyle

486.62\sin{13.2^o} = 111.17
\)

then \(\displaystyle 111.17 - 42 = 69.11m \)(the hieght of the hill)