# Thread: Please Help, Horizontal Lengths!!!

1. ## Please Help, Horizontal Lengths!!!

Hi all,

I'm new here so big hi to everyone.

O.k. I have an assignment question here, its the last of 5 and i'm chuffed because i've solved all the rest but now my heads a bit broken.

this is the last question:

"three lines on a site form a triangle on plan. the HORIZONTAL lengths were measured and found to be 20.988, 38.011 and 25.101.

Determine the angles etc."

I can answer the question no prob, but i'm being cautious here, the words horizontal indicates that these are not the lengths of the lines of the trianlge. they may well be and i'm just being paranoid but this seems a trick question to me.

so my question is, can i find the length of an angled line from the horizontal length alone?

I'm sure its impossible because i have no angles, with the angle its easy, additionally, i have no vertical component in order to find the angles using tan..........im lost.

so guys, any thoughts or am i just reading to much into the question?

2. Originally Posted by Parton-Bill
Hi all,

I'm new here so big hi to everyone.

O.k. I have an assignment question here, its the last of 5 and i'm chuffed because i've solved all the rest but now my heads a bit broken.

this is the last question:

"three lines on a site form a triangle on plan. the HORIZONTAL lengths were measured and found to be 20.988, 38.011 and 25.101.

Determine the angles etc."

I can answer the question no prob, but i'm being cautious here, the words horizontal indicates that these are not the lengths of the lines of the trianlge. they may well be and i'm just being paranoid but this seems a trick question to me.

so my question is, can i find the length of an angled line from the horizontal length alone?

I'm sure its impossible because i have no angles, with the angle its easy, additionally, i have no vertical component in order to find the angles using tan..........im lost.

so guys, any thoughts or am i just reading to much into the question?

Yes a definate trick question has been posed,
I cant see a feasible way of computing the angles of the triangle unless the height and or classification of the triangle is known,

if the horizontal lengths simply represent the actual lengths of the triangle and by imposing that we know nothing about the shape of the triangle and thus assume to be scalene then to solve the angles, simply employ the cosine rule.

Hope this helps,

David

3. Thanks David,

thanks very much for that matey, i came to the same conclusion.

after trying everything else i still feel its a trick question but with no other information Ive decided to use cosine.

many thanks

4. Hello, Parton-Bill!

The term "horizontal" was misused (and highly misleading) here.
I think those three lengths are simply the sides of the triangle. .**

A triangle has sides: . $a= 20.988,\;\;b= 38.011,\;\;c= 25.101$

Determine the angles of the triangle.
Code:
                  B
*
*  *
25.101   *     * 20.998
*        *
*           *
*              *
A *  *  *  *  *  *  * C
38.011
Use the Law of Cosines . . .

$\cos A \;=\;\frac{b^2+c^2-a^2}{2bc} \;=\;\frac{38.001^2 + 25.101^2-20.998^2}{2(38.011)(25.101)} \;=\;0.856281392 \quad\Rightarrow$ . . $A \;\approx\;31.1^o$

$\cos B \;=\;\frac{a^2+c^2-b^2}{2ac} \;=\;\frac{20.998^2 + 25.101^2-38.011^2}{2(20.998)(25.101)} \;=\;\text{-}0.354658159 \quad\Rightarrow$ . $B \;\approx\;110.8^o$

$\cos C \;=\;\frac{a^2+b^2-c^2}{2ab} \;=\;\frac{20.998^2 + 38.011^2 - 25.101^2}{2(20.998)(38.011)} \;=\;0.78662162 \quad\Rightarrow$ . . . $C \;\approx\;38.1^o$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

If those were truly horizontal distances,
. . then $\Delta ABC$ is "projected" onto a horizontal line.
Code:
                B
*
*: *
* :   *
*  :     *
*   :       * C
*    :   *   :
*     *       :
*  *   :       :
A *       :       :
:       :       :
- - * - - - * - - - * - -
A'      B'      C'

And we see that: . $A'B' + B'C\:\!'\:\text{ must equal }\:A'C\:\!'$
. . . which is not true of the given segments.