# Math Help - trigonomerty - modelling and problem solving

1. ## trigonomerty - modelling and problem solving

Three points A, B and C are on a horizontal line such that AB = 70m and BC = 35m. The angles of elevation of the top of the tower are x, y and z, where tan x = 1/13, tan y = 1/15, and tan z = 1/20. The foot of the tower is at the same level as A, B and C. Find the height of the tower.

I can draw the diagram and have 3 equations in 2 unkowns, can't go any further. Please help

2. ## Re: trigonomerty - modelling and problem solving

I have tried out the three equations. the value of h i got was 35 but on verification with third equation i got tan z = 1/16. Please verify the values.

3. ## Re: trigonomerty - modelling and problem solving

Can you explain further. Thanks

4. ## Re: trigonomerty - modelling and problem solving

Hello, elmidge!

I agree with ibdutt: . $\tan z$ must be $\tfrac{1}{16}$
But then, we don't need point $C$ or angle $z.$

$\text{Three points }A, B, C\text{ are on a horizontal line}$
. . $\text{ such that }AB = 70m\text{ and }BC = 35m.$

$\text{The angles of elevation of the top of the tower are }x, y, z,$
. . $\text{ where: }\,\tan x = \tfrac{1}{13},\;\tan y = \tfrac{1}{15},\text{ and }\tan z = \tfrac{1}{16}$

$\text{The foot of the tower is at the same level as }A, B\text{ and }C.$

$\text{Find the height of the tower.}$

The tower is: $h = PQ.$
Code:
    P *
|** *
| * *   *
|  *  *     *
h |   *   *       *
|    *    *         *
|     *     *           *
|    x *    y *           z *
Q * - - - * - - - * - - - - - - - *
13h   A  70   B      35       C
In $\Delta PQA: \:\tan x = \frac{h}{QA} \quad\Rightarrow\quad \frac{1}{13} = \frac{h}{QA} \quad\Rightarrow\quad QA = 13h$

In $\Delta PQB: \:\tan y = \frac{h}{13h+70} = \frac{1}{15} \quad\Rightarrow\quad 15h = 13h +70$

Therefore: . $h \:=\:35\text{ m}$

5. ## Re: trigonomerty - modelling and problem solving

You're both making the assumption that the base of the tower is on the same line as A, B and C.
That needn't be the case, and if it isn't it's perfectly possible to arrive at an answer consistent with the given values.
I have to go out now, but I can post a solution tomorrow unless either of you do so in the meanwhile.

6. ## Re: trigonomerty - modelling and problem solving

Let the tower be of height h with base at D, (not on the line ABC), and top E.
Then,

$\frac{DE}{AD}=\frac{1}{13},$ so $AD = 13h,$ and similarly $BD = 15h \text{ and } CD = 20h.$

We now have a triangle $ACD,$ with $AC =105, AD = 13h, CD = 20h,$ with a point

$B \text{ on } AC \text{ such that } AB = 70, BC = 35, \text{ and } BD = 15h.$

Here are two different methods for finding the value of h.

(1) Using the cosine rule in the two triangles ABD and ACD.

$\cos A = \frac{169h^{2}+70^{2}-225h^{2}}{2.13h.70} \text{ from ABD, and }\cos A = \frac{169h^{2} + 105^{2}-400h^{2}}{2.13h.105} \text{ from ACD}.$

Tidy up and equate the two and find that $h^{2}=25, \text{ so, } h=5.$

(2) Using some co-ordinate geometry.

Arrange for the line ABC to be along the x-axis with A at the origin.

Take circles radii 13h, 15h and 20h at centres A, B and C respectively. The base of the tower D, will be at a common point of intersection.

The circles have equations

$x^{2}+y^{2}= 169h^{2},\dots (1),$

$(x-70)^{2}+y^{2}=225h^{2}, \dots (2)$

$(x-105)^{2}+y^{2}=400h^{2}, \dots (3).$

Expand the two brackets and subtract (1) from (2), (1) from (3) yielding (4) and (5),

$-140x+4900=56h^{2}, \dots (4)$

$-210x+11025 = 231h^{2} \dots (5).$

Now multiply (4) by 3, (5) by 2 and subtract one from the other to get rid of x.

Tidy up and again find that $h^{2}=25, \text{ so, } h=5.$

7. ## Re: trigonomerty - modelling and problem solving

I found this weblog rather helpful. The particulars and exact recommendation are specifically what I was wanting. I’ve book marked and will definitely be returning. all newspaper adsThanks for the information in this blog.The posting in this site is very cool and also interesting.I had read the entire blog and I came to know many things which I don't know before. upass4sure certificationI am sure that the visitors who visit this site will also be enjoying reading the posts.Keep it up.Waiting for new posts from you to be posted in this site.

Thank you.