# Math Help - bearings.. help

1. ## bearings.. help

Height of a tower
From a point P on level ground, the angle of elevation of the top of a tower is 26 °50.' From a point 25meters closer to the tower and on the same line with P and the base of the tower, the angle of elevation of the top is 53°30.' Approximate the height of the tower.

_______________________
25m

tan 26°50'=height (OPPOSITE)
25m+x

tan53° 30'=h
x

Generalize to the case where the first angle is alpha, the second angle is beta, and the distance between the two points is d. Express the height h of the tower in terms of d, alpha and beta.

ά=tan inverse htx
25 meters

β=90-°ά

2. Originally Posted by cocoknny
Generalize to the case where the first angle is alpha, the second angle is beta, and the distance between the two points is d. Express the height h of the tower in terms of d, alpha and beta.

ά=tan inverse htx
25 meters

β=90-°ά
Let the distance between point P and the tower be x. Let the height of the tower be y.

Then
$tan(\alpha) = \frac{y}{x}$
and
$tan(\beta) = \frac{y}{x - d}$

We need to solve for y, given $\alpha$, $\beta$, and d.

So from the first equation:
$x = \frac{y}{tan(\alpha)}$

Inserting this value of x into the second equation gives:
$tan(\beta) = \frac{y}{\frac{y}{tan(\alpha)} - d}$

$tan(\beta) = \frac{y}{\frac{y}{tan(\alpha)} - d} \cdot \frac{tan(\alpha)}{tan(\alpha)}$

$tan(\beta) = \frac{y~tan(\alpha)}{y - d~tan(\alpha)}$

Can you finish it from here?

-Dan

3. Originally Posted by cocoknny
y___
y-d
What is this?

Originally Posted by cocoknny
is my top part right?
I don't know because I don't know what "htx" is supposed to mean.

-Dan

4. tan 26°50'=h
25m+x

5. ## hello

Oh don't worry about that part. h suppose to be the height

6. Originally Posted by cocoknny
tan 26°50'=h
25m+x
It looks good if you have defined x as the distance from the second point to the tower.

That's one equation. You need to use the other to eliminate x.

-Dan

7. ## hello

YUP that's what i did. Thanks hun