# Math Help - trigonomerty - modelling and problem solving

1. ## trigonomerty - modelling and problem solving

A cat sitting on the edge of a straight river bank spots a bird sitting in a tree directly across the river and on the river's edge. The angle of elevation from the cat to the bird is 15 degrees. The cat then moves 25m along the river bank, and now spots the same bird at an angle elevation of 13 degrees. How high is the tree?

I have tried to solve this using simultaneous equations and get an answer of 42m which is wrong. Can you help please??

2. ## Re: trigonomerty - modelling and problem solving

Hello, elmidge!

This is a 3-D problem . . . very hard to draw.

A cat sitting on the edge of a straight river bank spots a bird sitting in a tree directly across the river
and on the river's edge. .The angle of elevation from the cat to the bird is 15o.
The cat then moves 25m along the river bank, and the angle elevation of the bird is 13i.
How high is the tree?

This is the diagram with the first position of the cat and the tree.
Code:
                        * B
*  |
* 75o |
*        | h
*           |
* 15o          |
C1 * - - - - - - - - * T
h tan75o
The cat is at $C_1$.
The bird is at $B$.
The tree is: $h = BT.$
$\angle C_1 = 15^o \quad\Rightarrow\quad \angle B = 75^o$
Hence: . $C_1T = h\tan75^o$

This is the diagram with the second position of the cat and the tree.
Code:
    B *
|   *
| 77o   *
h |           *
|               *
|               13o *
T * - - - - - - - - - - - * C2
h tan77o
The cat is at $C_2.$
The bird is at $B.$
The tree is: $h = BT.$
$\angle C_2 = 13^o \quad\Rightarrow\quad \angle B = 77^o$
Hence:. $C_2T \,=\,h\tan75^o$

Looking down at the ground, this is the diagram.
Code:
          T *
|  *
|     *
h tan75o |        *  h tan77o
|           *
|              *
C1 * - - - - - - - - * C2
25
Pythagorus: . $(h\tan75^o)^2 + 25^2 \:=\:(h\tan77^o)^2$

. . . . . . . . . . $h^2\tan^275^o + 625 \:=\:h^2\tan^277^o$

. . . . . $h^2\tan^277^o - h^2\tan^275^o \:=\:625$

. . . . . . $h^2(\tan^277^o - \tan^275^o) \:=\:625$

. . . . . . . . . . . . . . . . . . . . $h^2 \:=\:\frac{625}{\tan^277^o - \tan^275^o}$

. . . . . . . . . . . . . . . . . . . . . $h \:=\:\sqrt{\frac{625}{\tan^277^o - \tan^275^o}}$

Therefore: . $h \;\approx\;11.37\text{ m}$

3. ## Re: trigonomerty - modelling and problem solving

We could also do it as indicated .