# Plz solve these questions.(Trignometry statement questions)

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• Jul 14th 2008, 07:59 PM
prashantvrm
Plz solve these questions.(Trignometry statement questions)
• Jul 14th 2008, 09:57 PM
Jhevon
Quote:
you seem to be missing information from number 2), i think you meant to state the angle of elevation, but didn't. anyway, we can still do this problem in principle. the other problems are similar, try attacking them in the same way. that is, draw diagrams and figure out what trig ratio you need to solve for the unknown. i would hope you know what is meant by "angle of elevation" and "angle of depression."

anyway, in this problem, let the angle of elevation be $\displaystyle x$ and the distance from the foot of the pole to the point be $\displaystyle d$ (see the diagram below).

we can use the tangent trig ratio here.

problem 2:

$\displaystyle \tan x = \frac {\text{opposite}}{\text{adjacent}} = \frac {12}d$

$\displaystyle \Rightarrow d = \frac {12}{\tan x}$

now, if we know what x is, we can find the distance. otherwise, leave it like that. try the others, be sure to draw diagrams so you know how to think about the problem
• Jul 15th 2008, 04:29 AM
Soroban
Hello, prashantvrm!

Quote:

4. The angle of elevation to the top of a chimney from a point on the ground is 30°.
After walking 50 m towards the chimney, the angle of elevation becomes 45°.
Find the height of the chimney.

Code:

                                  * C                               * * |                           *  *  |                       *    *    | h                   *      *      |               *        *        |           * 30°      * 45°      |       * - - - - - - * - - - - - - *       A    50      B      x      D
Let $\displaystyle CD = h$, the height of the chimney.

When the observer stands at $\displaystyle A,\:\angle CAD = 30^o.$

At $\displaystyle B$ (where $\displaystyle AB = 50$), $\displaystyle \angle CBD = 45^o.$

Let $\displaystyle x = BD.$

In right triangle $\displaystyle CDA\!:\;\frac{h}{x+50} \:=\:\tan30^o \:=\:\frac{1}{\sqrt{3}} \quad\Rightarrow\quad \sqrt{3}h \:=\:x + 50\;\;{\color{blue}[1]}$

In right triangle $\displaystyle CDB\!:\;\frac{h}{x} \:=\:\tan45^o \:=\:1 \quad\Rightarrow\quad x \:=\:h\;\;{\color{blue}[2]}$

Substitite [2] into [1]: .$\displaystyle \sqrt{3}h \:=\:h + 50 \quad\Rightarrow\quad \sqrt{3}h - h \:=\:50$

Factor: .$\displaystyle (\sqrt{3}-1)h \:=\:50 \quad\Rightarrow\quad h \:=\:\frac{50}{\sqrt{3}-1} \;\approx\;\boxed{68.3\text{ m}}$

• Jul 15th 2008, 04:50 AM
Soroban
Hello again, prashantvrm!

Quote:

3. The top of a tree is broken and the broken part makes a 30° angle with the ground.
The distance from the tip of the tree to the base of the tree is 40 m.
Find the height of the tree.

Code:

    A *       |  *       |      *       |          *       |          30° *     C * - - - - - - - - - * B               40
This is a 30-60 right triangle.

We know the ratios of ths sides . . . .$\displaystyle AC : BC : AB \:=\:1 : \sqrt{3} : 2$

Hence, we have: .$\displaystyle AC : BC : AB \;=\;\frac{40\sqrt{3}}{3} : 40 : \frac{80\sqrt{3}}{3}$

The height of the tree is: .$\displaystyle AC + AB \;=\;\frac{40\sqrt{3}}{3} + \frac{80\sqrt{3}}{3} \;=\;\frac{120\sqrt{3}}{3} \;=\;\boxed{40\sqrt{3}\text{ m}}$

• Jul 19th 2008, 07:18 PM
aronfelizardo
hai
just find the distance and u can find the answer .....
its just too easy.......
(Cool)