# Math Help - last one

1. ## last one

Ok, heres the last one i need help with

From a window 4.2m above the ground the angle of depression ot the foot of the building across the road is 24(degrees) and the angel of elevation of the to pof the same building is 34(dg) Determine correct to the nearest cm the width of the road and the height of the building.

Many thanks to everyone for their help

2. Originally Posted by damonneedshelp
Ok, heres the last one i need help with

From a window 4.2m above the ground the angle of depression ot the foot of the building across the road is 24(degrees) and the angel of elevation of the to pof the same building is 34(dg) Determine correct to the nearest cm the width of the road and the height of the building.

Many thanks to everyone for their help
You think you can draw the figure on paper?

I assume you can, by yourself..

In the lower right triangle,
If w = width of road,
tan(24deg) = 4.2/w
So, w = 4.2/tan(24deg) = 9.4333 m = 9.43 m -----answer.

In the upper right triangle,
If y = height of the opposite building above the 4.2m,
tan(34deg) = y/w = y/9.43
So, y = (9.43)(tan(34deg)) = 6.36 m.
Therefore,
the total height of the opposite building is 4.2 +6.36 = 10.56 m ---aqnswer.

3. Hello, Damon!

From a window 4.2m above the ground,
the angle of depression ot the foot of the building across the road is 24°
the angel of elevation of the top of the same building is 34°.
Determine to the nearest cm the width of the road and the height of the building.

Here's the diagram for ticbol's solution.
Code:
                              * C
*  |
*     |
*        |
*           |y
*              |
*                 |
* 34°     w          |
A * - - - - - - - - - - - + D
|   * 24°               |
|       *               |
4.2 |           *           | 4.2
|               *       |
|                   *   |
B +-----------------------* E
: - - - -  w  - - - - - :

The window is at $A$, 4.2 m above the ground.
. . $AB\,=\,DE\,=\,4.2$

Let $y \,=\,CD$ and $w \,=\,BE\,=\,AD.$

In right triangle $ADE\!:$
. . $\tan24^o \:=\:\frac{4.2}{w}\quad\Rightarrow\quad w \:=\:\frac{4.2}{\tan24^o}\quad\Rightarrow\quad \boxed{w \:\approx\:9.43\text{ m}}$

In right triangle $CDE:$
. . $\tan34^o \:=\:\frac{y}{w}\quad\Rightarrow\quad y \:=\:9.43\tan34^o\quad\Rightarrow\quad y \:\approx\:6.36\text{ cm}$

The height of the building is about: . $6.36 + 4.20 \:=\:\boxed{10.56\text{ m}}$