# Math Help - cosine law - word problem

1. ## cosine law - word problem

1. A helicopter hovers directly above the landing pad on the roof of a 125 m high building. A person is standing 145m from the base of the building. the angle of elevation of the helicopter from this person is 58 degrees. assume the landing pad is at the edge of the roof closest to the person on the ground. how high is the helicopter hovering about the landing pad?

2. One ship leaves a port and sails at 17 km/h on a bearing of 024 degrees. A second ship leaves the same port at the same time and sails at 21 km/h on a bearing of 071 degrees. How far apart are the 2 ships after 2 h?
(mainly i dont know how to draw a diagram for it)

2. Originally Posted by checkmarks
1. A helicopter hovers directly above the landing pad on the roof of a 125 m high building. A person is standing 145m from the base of the building. the angle of elevation of the helicopter from this person is 58 degrees. assume the landing pad is at the edge of the roof closest to the person on the ground. how high is the helicopter hovering about the landing pad?
First you need a diagram, so see the attachment.

Now we need to ignore the eye height of the observer. Then we see from the
diagram that:

$\tan(58)=\frac{125+x}{145}$,

where $x$ is the height of the helicopter above the pad.
Evaluating the tan and simplifying gives:

$1.60 \times 145=125+x$,

or:

$x=1.60 \times 145-125=107 \mbox{ m}$

RonL

3. Hello, checkmarks!

]1. A helicopter hovers directly above the landing pad on the roof of a 125 m high building.
A person is standing 145 m from the base of the building.
The angle of elevation of the helicopter from this person is 58°.
Code:
                          *H
* |
*   | y
*     |
*       *A
*         |
*           |
*             | 125
*               |
* 58°             |
O* - - - - - - - - - *B
145
This one doesn't require the Law of Cosines.

The observer is at $O$. .The helicopter is at $H$.
The building is $AB = 125$. . $OB = 145$.
Let $y = HA$

In right triangle $HBO$, we have: . $\tan58^o \:=\:\frac{y + 125}{145}$

Therefore: . $y \;=\;145\tan58^o - 125 \;\approx\;107$ m.

2. One ship leaves a port and sails at 17 km/h on a bearing of 024 degrees.
A second ship leaves the same port at the same time and sails at 21 km/h
on a bearing of 071 degrees.
How far apart are the two ships after 2 hours?

The bearing is measured clockwise from North.
After two hours, the positions of the ships look like this:
Code:
                  A
:           *
:          *  *
:         *     *  x
:        *        *
:       *           *
:      *34            *
:     *                 * B
:24° *              *
:   *           *
:  * 47°    * 42
: *     *
:*  *
*
P
Now we can use the Law of Cosines.

$x^2\;=\;34^2 + 42^2 - 2(34)(42)\cos47^o \:=\:972.2126837$

Therefore: . $x \;\approx\;31.2$ km.

4. Originally Posted by checkmarks

2. One ship leaves a port and sails at 17 km/h on a bearing of 024 degrees. A second ship leaves the same port at the same time and sails at 21 km/h on a bearing of 071 degrees. How far apart are the 2 ships after 2 h?
(mainly i dont know how to draw a diagram for it)