# precalc help (law of sines and cosines)

• Mar 22nd 2009, 09:49 AM
jul07010
precalc help (law of sines and cosines)
hard to explain without a diagram, but:
a point on the ground is 200 feet from a water tower
the angle of elevation to the top of the tower is 18 degrees
the angle of elevation to the bottom of the tower is 15 degrees
how tall is the water tower?
(i think the water tower is on a hill?)
thanks!
• Mar 22nd 2009, 12:37 PM
u2_wa
Quote:

Originally Posted by jul07010
hard to explain without a diagram, but:
a point on the ground is 200 feet from a water tower
the angle of elevation to the top of the tower is 18 degrees
the angle of elevation to the bottom of the tower is 15 degrees
how tall is the water tower?
(i think the water tower is on a hill?)
thanks!

Hello jul07010:

I also think water tower is on the hill

$base=200m$

$tan(18)=\frac{Height \color{red}of\color{black} hill+ tower}{200}
$

$\frac{tan(18)}{200}=$Height of hill+ tower

Height of hill= $\frac{tan(15)}{200}$

Height of tower= (Height of hill+ tower)-(Height of hill)
• Mar 22nd 2009, 03:17 PM
Soroban
Hello, jul07010!

I agree with u2_wa: we don't need the Law of Sines . . .

Quote:

A point on the ground is 200 feet from a water tower.
The angle of elevation to the top of the tower is 18°.
The angle of elevation to the bottom of the tower is 15°.
How tall is the water tower?
(i think the water tower is on a hill?) yes!

Code:

    B *       | *       |  *     h |    *       |      *       |        *     C *          *       |  *        *       |      *      *     y |    200  *  3° *       |              *  *       |              15° * *     D * - - - - - - - - - - - * A                   x

The height of the tower is: $h = BC$
Let $y = CD,\;x = DA.$

$AC$ is the ground: $AC = 200.$
$\angle CAD = 15^o,\;\angle BAD = 18^o$

In right triangle $CDA:\;\begin{array}{ccccccc}\sin15^o \:=\:\frac{y}{200} &\Rightarrow& y \:=\:200\sin15^o & {\color{blue}[1]}\\
\cos15^o \:=\:\frac{x}{200} & \Rightarrow & x \:=\:200\cos15^o & {\color{blue}[2]} \end{array}$

In right triangle $BDA\!:\;\tan18^o \:=\:\frac{h+y}{x} \quad\Rightarrow\quad h \;=\;x\tan18^o - y$ .[3]

Substitute [1] and [2] into [3]: . $h \;=\;(200\cos15^o)\tan18^o - 200\sin15^o$

Therefore: . $h \;\approx\;11$ ft.