# trigonomerty - modelling and problem solving

• Apr 6th 2013, 05:37 PM
elmidge
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??
• Apr 6th 2013, 07:03 PM
Soroban
Re: trigonomerty - modelling and problem solving
Hello, elmidge!

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

Quote:

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 $\displaystyle C_1$.
The bird is at $\displaystyle B$.
The tree is: $\displaystyle h = BT.$
$\displaystyle \angle C_1 = 15^o \quad\Rightarrow\quad \angle B = 75^o$
Hence: .$\displaystyle C_1T = h\tan75^o$

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 $\displaystyle C_2.$
The bird is at $\displaystyle B.$
The tree is: $\displaystyle h = BT.$
$\displaystyle \angle C_2 = 13^o \quad\Rightarrow\quad \angle B = 77^o$
Hence:.$\displaystyle C_2T \,=\,h\tan75^o$

Looking down at the ground, this is the diagram.
Code:

          T *             |  *             |    *   h tan75o |        *  h tan77o             |          *             |              *         C1 * - - - - - - - - * C2                     25
Pythagorus: .$\displaystyle (h\tan75^o)^2 + 25^2 \:=\:(h\tan77^o)^2$

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

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

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

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

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

Therefore: .$\displaystyle h \;\approx\;11.37\text{ m}$
• Apr 7th 2013, 05:25 AM
ibdutt
Re: trigonomerty - modelling and problem solving
We could also do it as indicated . Attachment 27846