# Plz solve this trignometry question.Tomorrow is my exam.

• Jul 20th 2008, 03:16 AM
prashantvrm
Plz solve this trignometry question.Tomorrow is my exam.
Tomorrow is my maths 1 exam and next day maths 2 exam so plz solve these question whole as soon as possible.

1)As observed from the top of a 100 m tall light house the angles of depression of two ships approaching it are $\displaystyle 30^o$ and $\displaystyle 45^o$. If one ship is directly behind the other, find the distance between the two sips.

2)The angles of elevation of the top of a rock at top and foot of a 100 m high tower are 30^o and 45^o respectively>Find the height of the rock.
• Jul 20th 2008, 04:57 AM
Soroban
Hello, prashantvrm!

They used "nice" angles in these problems.
We can solve them without a calculator.
. . I'll walk through the first one.

Quote:

1) As observed from the top of a 100 m lighthouse,
the angles of depression of two ships approaching it are 30° and 45°.
If one ship is directly behind the other, find the distance between the two ships.

Code:

    A * - - - - - - - - - - - C       | * * 30°       |  *  *       |    *    *   100 |      *      *       |        *        *       |      45° *      30° *       * - - - - - - * - - - - - - *       B      y      P      x      Q
The lighthouse is $\displaystyle AB = 100.$
The ships are at $\displaystyle P$ and $\displaystyle Q.$

$\displaystyle \angle CAQ = 30^o$, hence: $\displaystyle \angle AQB = 30^o$
$\displaystyle \angle CAP = 45^o$, hence: $\displaystyle \angle APB = 45^o$

Let $\displaystyle x = PQ,\;y = BP$

In $\displaystyle \Delta ABP\!:\;\tan45^o \:=\:\frac{y}{100} \quad\Rightarrow\quad\frac{y}{100} \:=\:1 \quad\Rightarrow\quad y \:=\:100$ .[1]

In $\displaystyle \Delta ABQ\!:\;\;\tan30^o \:=\:\frac{100}{x+y} \quad\Rightarrow \quad \frac{1}{\sqrt{3}} \:=\:\frac{100}{x+y} \quad\Rightarrow\quad x+y \:=\:100\sqrt{3}$ .[2]

Substitute [1] into [2]: .$\displaystyle x + 100 \:=\:100\sqrt{3}$

. . Therefore: .$\displaystyle x \;=\;100\sqrt{3} - 100 \;=\;100(\sqrt{3}-1)$