1. ## 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.

2. 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

$\frac{height1}{base1}=\frac{height2}{base2}\ \Rightarrow\ \frac{1.4}{1.2}=\frac{tree\ height}{15}$

3. can you send me a diagram of this question?

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

5. With only using SSS≈, SAS≈ and AA≈, how can you prove these 2 triangles are similar?

6. I hope this is ok, the tree is very tall in comparison to Sarah.

In a right-angled triangle, one of the 3 angles is 90 degrees.
Hence if 2 others are the same,
the 3rd and final angles are also the same and are $90^o-\theta$