# Thread: angles of elevation and depression

1. ## angles of elevation and depression

Could you please help me with this problem thnx.From a coastal lookout point A,80m above sea level,a man sights two boats B and C in the same direction.The angles of depression of the two boats are 13degrees and 24degrees respectively.Find the distances of B and C from the point below A and so find find the distance between the boats.thanks againsalahu

2. Hello, salahu!

From a coastal lookout point A, 80m above sea level,
a man sights two boats B and C in the same direction.
he angles of depression of the two boats are 13° and 24°. respectively.
Find the distances of B and C from the point below A
and so find find the distance between the boats.
Code:
    A * - - - - - - - - - - - - - - E
|  *  *  13°
|     *     *
80 |        *        *
|           *           *
|          24° *         13°  *
* - - - - - - - - * - - - - - - - - *
D       x         C       y         B

$\angle EAB = 13^o.\;,\text{ Hence, }\angle ABD = 13^o$

$\angle EAC = 24^o.\;\text{ Hence, }\angle ACD = 24^o$

Let $x \:=\:DC.\;y \:=\:CB$

In right triangle $ADC\!:\;\;\tan24^o \;=\;\frac{80}{x}\quad\Rightarrow\quad x \;=\;\frac{80}{\tan24^o}$ .[1]

In right triangle $ADC\!:\;\;\tan13^o \;=\;\frac{80}{x+y}\quad\Rightarrow\quad x + y \;=\;\frac{80}{\tan13^o}$ .[2]

Substitute [1] into [2]: . $\frac{80}{\tan24^o} + y \;=\;\frac{80}{\tan24^o}$

Therefore: . $y \;=\;\frac{80}{\tan13^o} - \frac{80}{\tan24^o} \;=\;166.835128$ m.