# Similar Triangles help

• May 3rd 2010, 02:47 PM
sinjid9
Similar Triangles help
To find the height of a tree, Sarah placed a mirror on the ground 15 m from the base of a tree. She walked backward until she could see the top of the tree in the centre of the mirror. At that position she was 1.2m from the mirror and her eyes were 1.4m from the ground.
• May 3rd 2010, 02:58 PM
Quote:

Originally Posted by sinjid9
To find the height of a tree, Sarah placed a mirror on the ground 15 m from the base of a tree. She walked backward until she could see the top of the tree in the centre of the mirror. At that position she was 1.2m from the mirror and her eyes were 1.4m from the ground.

Hi sinjid9,

The triangle from the mirror to the tree to the base of the tree
is a magnified version of the triangle from the mirror to Sarah's eyes to her feet,
if the tree is taller than her, which the dimensions confirm.

You may write

$tan\theta=\frac{h}{15}=\frac{1.4}{1.2}$

multiply both sides by 15 to find h

Or

$\frac{height1}{base1}=\frac{height2}{base2}\ \Rightarrow\ \frac{1.4}{1.2}=\frac{tree\ height}{15}$
• May 3rd 2010, 04:04 PM
sinjid9
can you send me a diagram of this question?
• May 3rd 2010, 04:10 PM
Soroban
Hello, sinjid9!

Quote:

To find the height of a tree, Sarah placed a mirror on the ground 15 m from the base of a tree.
She walked backward until she could see the top of the tree in the centre of the mirror.
At that position she was 1.2m from the mirror and her eyes were 1.4m from the ground.

Code:

                              o C                             * |                           *  |                         *    |     A o              *      | h       | *          *        |   1.4 |  *      *          |       |  θ *  * θ          |     B o - - - o - - - - - - - o D         1.2  M      15

Sarah is $AB = 1.4$
The tree is $CD = h.$
The mirror is at $M\!:\;\;BM = 1.2,\;MD = 15$

Since "angle of reflection equals angle of incidence",
. . we have: . $\angle AMB \,=\,\angle CMD \,=\,\theta.$

Hence: . $\Delta CDM \sim \Delta ABM$

Therefore: . $\frac{h}{15} \;=\;\frac{1.4}{1.2} \quad\hdots\;\text{etc.}$

• May 3rd 2010, 04:14 PM
sinjid9
With only using SSS≈, SAS≈ and AA≈, how can you prove these 2 triangles are similar?
• May 3rd 2010, 04:24 PM
the 3rd and final angles are also the same and are $90^o-\theta$