My Math Forum slope and inclination

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 April 19th, 2017, 11:47 PM #1 Member   Joined: Nov 2012 From: The Netherlands Posts: 30 Thanks: 1 slope and inclination Hello All This is on behalf of my 14 year old son. He came to me with this issue and I come here http://s10.postimg.org/qumivopft/Bridge.png Is it possible, someone can assist me , assisting my son here ? Thank you Regards *PS: the word 'hoogte' in the picture means 'height' (in dutch)
 April 20th, 2017, 12:18 AM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,167 Thanks: 383 Math Focus: Yet to find out. a) $tan\theta = \dfrac{O}{A} = 0.075$ $\therefore \theta = tan^{-1} (0.075) \approx 4.289$. b) $tan(4.289) = \dfrac{O}{60}$ O is the height of the pillar. $O = 60tan(4.289) \approx 4.5m$ EDIT: I forgot c). Probably many ways, but using cosine rule would work. $c^2 = a^2 + 2ab cos(C)$ $c = BC = 20m, \ a = b = \sqrt{(4.5-0.04)^2 + 20^2}$ Then solve for the angle C. Thanks from hello_math Last edited by Joppy; April 20th, 2017 at 01:10 AM.
April 20th, 2017, 01:40 AM   #3
Senior Member

Joined: Feb 2016
From: Australia

Posts: 1,167
Thanks: 383

Math Focus: Yet to find out.
Quote:
 Originally Posted by Joppy $c^2 = a^2 + b^2 - 2ab cos(C)$ Then solve for the angle C.
Whoops. Sorry. This is not the way to go about it anyway.

The triangle formed is isosceles, and the angle can be solved in the following way,

$tan\frac{\theta}{2} = \dfrac{10}{4.5 - 0.045}$

$\theta = 2tan^{-1} (\dfrac{10}{4.5 - 0.045}) \approx 132^{\circ}$

 April 20th, 2017, 01:43 AM #4 Member   Joined: Nov 2012 From: The Netherlands Posts: 30 Thanks: 1 Yes ..I was wondering about 'c' , and didnt know how to get that through Thank You SO Much ! Edit: $tan\frac{\theta}{2} = \dfrac{10}{4.5 - 0.045}$ $\theta = 2tan^{-1} (\dfrac{10}{4.5 - 0.045}) \approx 132^{\circ}$ can you please show this in 'notation' (like in the 1st post) ? Last edited by hello_math; April 20th, 2017 at 01:45 AM.
 April 20th, 2017, 01:47 AM #5 Member   Joined: Nov 2012 From: The Netherlands Posts: 30 Thanks: 1 our messages crossed each other I am absolutely Fine now ! Thank You
 April 20th, 2017, 12:29 PM #6 Member   Joined: Nov 2012 From: The Netherlands Posts: 30 Thanks: 1 Hi I wanted to know a bit more on 'c' as I was not fully clear on it myself I have made the figure here [https://s9.postimg.org/rm1t19dnz/20-...7_22-18-09.png ] , with a point S added. As I see: angle S = 90 degrees tan theta = perpendicular/base thus tan theta = BS/ FS tan theta = 20/ .04 tan theta = 500 but I am not able to interpret your way of : http://s29.postimg.org/7d29k5enr/20_...7_22_18_09.png could you explain the 'C' part a bit more please. Thank You Last edited by hello_math; April 20th, 2017 at 12:45 PM.
 April 20th, 2017, 04:28 PM #7 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,167 Thanks: 383 Math Focus: Yet to find out. I would first like to apologise for the previous posts. They're sloppy and cryptic at best. For finding the angle, there are several methods. One of which, involves applying the codeine rule to the triangle BFC. In this case, we know the length opposite the angle. In your first diagram, this length would be S + S = 40m. In my use of the codeine rule, it is the lower case 'c'. But we also know some more information. We know that the lengths BF and CF are equal. Let's look at BF on the diagram. To find its length, we could construct a little triangle whose hypotenuse is length BF, it's height is (4.5m - 0.045m), and its base is 20m. And since BF = CF, we just need to calculate, $CF = BF = \sqrt{(4.5 - 0.045)^2 + 20^2}$. Now we have the cos rule, $BC^2 = CF^2 + BF^2 - 2*BF*CF* cos(\phi)$. I've called the obtuse angle we are trying to find here '$\phi$'. The above equations has one unknown, and 3 knowns, so we can solve it! You should find that $\phi = 154.9^{\circ}$. Note that this is not the same answer as the other method due to my mistake! I haven't the faintest clue what I was thinking. For the other method, we turn the big triangle BCF into two smaller triangle as you have done. Then our big angle (the one we are trying to find) gets cut in half! So if we try to find this 'half angle', we get, $tan \left( \dfrac{\phi}{2} \right)$ = $\dfrac{\mathbb{20}}{4.5 - 0.04}$. Solving this one should net you the same answer as before. Note I've used $\phi$ as in the other method to denote the angle we are trying to find. Hopefully this makes more sense, and sorry to cause confusion with the other posts. Thanks from hello_math Last edited by Joppy; April 20th, 2017 at 04:35 PM. Reason: Rendering issues

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