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I've been trying to brush up on my trigonometry, but I have realised that I need some help. I'm in the process of building a table, where the legs are connected to the table at a 6 degree angle in both the x and y axis if looking from the bottom of the table.

The dimensions of the table is fixed at (1) long edge = 150 cm, (2) short edge = 60 cm, (3) hight of table should be = 45 cm

I have three main questions. (1) How far from the table edges should the drill hole be, for the end of the table leg (the part hitting the floor) to be perpendicular(?) to the table edge. (2) how long do the table leg need to be, to get a table hight of 45 cm when angled at 6 degrees (3) If the table leg is angled 6 degrees towards the short and and the long end. What is the angle towards the corner of the table?

I've tried below. But I think Im doing something, wrong as the results dont make sense... Hope you guys can spot my errors.

My take on question 1 and 2

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Hint: What you have is a square pyramid with square length $a$ and height $45\,$cm with its apex perpendicularly above one of the squares vertex. Thus, two of the pyramids sides are right triangles with legs $45\,$cm and $a$, one of the angles being $6^{\circ}$, thus use tangens to calculate $a$. Finally, to get the length of the table leg, consider right triangle with one leg being the diagonal of the square (thus, having length $a\sqrt 2$) and the other being the height of the pyramid (thus, having length $45\,$cm). Hypotenuse of that triangle is table leg. Now use whatever trigonometric function you want to calculate the angle you want.

Alternatively, think of right square prism with square sides having length $a$ and its height $45\,$cm. Calculations are as described above.

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  • $\begingroup$ So basically the a is 4.73 cm (distance from table edge) and hight of table leg needs to be 45.25 cm to ensure a table height of 45 cm. $\endgroup$ – Chris_1983_Norway Jun 11 '17 at 10:00
  • $\begingroup$ So if I think of this as a cone. Basically if the table leg is fixed to the underside of the table 4.73 cm from the short edge and the long edge. Then the angle towards the corner point should be 6 degrees. This will also ensure there is 6 degrees from the table leg to the long edge and 6 degrees to the short edge? Is this correct? $\endgroup$ – Chris_1983_Norway Jun 11 '17 at 10:05
  • $\begingroup$ @Chris_1983_Norway, you are correct about lengths (up to rounding), but you are not correct about the angle. You can't have two right triangles with one leg congruent but not the other one to have congruent angles. One angle is $\arctan(a/45)$, while the other is $\arctan(a\sqrt 2/45)$. $\endgroup$ – Ennar Jun 11 '17 at 10:35
  • $\begingroup$ This is taking me into an unknown domain... would you be able to explain how this is calculated? $\endgroup$ – Chris_1983_Norway Jun 11 '17 at 11:27
  • $\begingroup$ So I have two right angled triangles from the hole in the table top to the edge. One to the long-edge and one to short-edge. I also have a right angled triangle from the same point of origin, but to the corner on the table top which is at the intersection of the long-edge and short-edge of the table. However I know none of the angles or lengths except the 90 degree angle. However knowing that the hole in the table should be 4.73 cm from both the long-edge and short-edge. I can calculate the hype of that triangle to 6.69 cm. $\endgroup$ – Chris_1983_Norway Jun 11 '17 at 11:36

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