last one

Printable View

January 10th 2007, 12:33 AM
damonneedshelp

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
January 10th 2007, 02:05 AM
ticbol

Quote:

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.
January 10th 2007, 03:23 AM
Soroban

Hello, Damon!

Quote:

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: . $6.36 + 4.20 \:=\:\boxed{10.56\text{ m}}$