Angle of elevation

[Shelby104]

Hello I'm doing a problem that is really bugging me I have it to a certain point the question is there's a mountain with an elevation of 17 degrees and 300 feet from the point the elevation is 32 degrees I need to find the height? So I no tan17=h/x and tan32=h/x-300 and Ive got my formula all the way to h[tan(32)/tan(17)-1]=300tan(32) but cannot figure this out please any help I really want to understand this problem because it seems hard so if I can get this I might be okay with my school thanks people

---/--------------/---------------|
--/--------------/----------------|
-/--------------/-----------------|h
/_17degrees /__32degrees___|
300ft------------------X


Tan17=h/x
Tan32=h/x-300

 

[pappus]

Hello I'm doing a problem that is really bugging me I have it to a certain point the question is there's a mountain with an elevation of 17 degrees and 300 feet from the point the elevation is 32 degrees I need to find the height? So I no tan17=h/x and tan32=h/x-300 and Ive got my formula all the way to h[tan(32)/tan(17)-1]=300tan(32) but cannot figure this out please any help I really want to understand this problem because it seems hard so if I can get this I might be okay with my school thanks people

/ / |
/ / |
/ / |h
/_17degrees /__32degrees___|
300ft X


Tan17=h/x
Tan32=h/x+300 <--- this is wrong. Compare your text above

1. You have 2 equations in 2 variables:
$\displaystyle \displaystyle{\tan(17^\circ)=\frac hx \\ \tan(32^\circ)=\frac h{x-300}}$

2. Determine h in the 1st equation: $\displaystyle \displaystyle{h = x \cdot \tan(17^\circ)}$ and replace h in the 2nd equation by this term. You'll get
$\displaystyle \displaystyle{\tan(32^\circ)=\frac{x \cdot \tan(17^\circ)}{x-300}}$

3. Solve for x. Afterwards use the value of x to determine the value of h.

 

[Shelby104]

I'm sorry I really don't understand how to find h if I don't have x or a height for 300 ft I'm very confused please help

 

[MarkFL]

Another approach:

You have:

$\displaystyle \tan(17^{\circ})=\dfrac{h}{x}$

$\displaystyle \tan(32^{\circ})=\dfrac{h}{x-300}$

This implies by solving both for x and equating:

$\displaystyle h\cot(17^{\circ})=h\cot(32^{\circ})+300$

Solve for h:

$\displaystyle h=\dfrac{300}{\cot(17^{\circ})-\cot(32^{\circ})}$

 

[Shelby104]

I understand but I don't have x I only have 300 feet Of x so how can I determine the rest without any info

 

[MarkFL]

Using the approach I gave, you don't need to know x. I solved both equations for x, then equated those two expressions which effectively eliminated x, leaving an equation in h alone, which is then solved in terms of the given data.

 

[soroban]

Hello Shelby104!

There is a mountain with an angle of elevation of 17 degrees from a certain point.
300 feet from that point the angle of elevation is 32 degrees.
Find the height of the mountain.

[SIZE=3]
                              * C
                           * *|
                        *   * |
                     *     *  |
                  *       *   | h
               *         *    |
            *           *     |
         * 17[SUP]o[/SUP]         * 32[SUP]o[/SUP]  |
      *---------------*-------* D
      A - -  300  - - B - x - :[/SIZE]

The height of the mountain is $\displaystyle h = CD.$

The first observation is made at $\displaystyle A.$
. . $\displaystyle \angle CAD = 17^o.$

The second observation is made at $\displaystyle B.$
. . $\displaystyle \angle CBD = 32^o.$
. . $\displaystyle AB = 300,\: BD = x.$

In $\displaystyle \Delta CDA\!:\;\tan17^o \,=\,\dfrac{h}{x+300} \quad\Rightarrow\quad x \:=\:\dfrac{h-300\tan17^o}{\tan17^o}$ .[1]

In $\displaystyle \Delta CDB\!:\;\tan32^o \,=\,\dfrac{h}{x} \quad\Rightarrow\quad x \:=\:\dfrac{h}{\tan32^o}$ .[2]

Equate [1] and [2]: .$\displaystyle \dfrac{h-300\tan17^o}{\tan17^o} \;=\;\dfrac{h}{\tan32^o}$

. . . . .$\displaystyle h\tan32^o - 300 \tan17^o\tan32^o \:=\:h\tan17^o$

. . . . . . . . . . . $\displaystyle h\tan32^o - h\tan17^o \;=\;300\tan17^o\tan32^o$

. . . . . . . . . . . $\displaystyle h(\tan32^o - \tan17^o) \;=\;200\tan17^o\tan32^o$

. . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle h \;=\;\dfrac{300\tan17^o\tan32^o}{\tan32^o-\tan17^o}$

. . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle h \;\approx\;179.6\text{ ft.}$