| Forum Home | ||||
| PC World Chat | ||||
| Thread ID: 129480 | 2013-02-25 03:21:00 | Geometry question | Tony (4941) | PC World Chat |
| Post ID | Timestamp | Content | User | ||
| 1329792 | 2013-02-25 03:21:00 | OK, I've just realized I've forgotten all my school geometry. I want to draw a symmetrical trapezium. The two parallel sides will be 310 mm and 180 mm. The other sides will be each be 280 mm. I need to calculate the distance between the two parallel sides or one of the angles. Any help greatly appreciated. |
Tony (4941) | ||
| 1329793 | 2013-02-25 03:29:00 | Well then you'd have a right-angled triangle with a hypotenuse of 280mm, and an adjacent side of 65mm. So cos(θ) = 65/280 gives you an angle between the longer parallel side and the non-parallel side of about 76.6* and a distance between parallel sides of about 272mm. I think I've done my maths right, I can't work out how to do inverse sine functions on windows calc :x | inphinity (7274) | ||
| 1329794 | 2013-02-25 03:35:00 | Where do you get 65 mm from? | Tony (4941) | ||
| 1329795 | 2013-02-25 03:53:00 | Where do you get 65 mm from?I get it - sorry. Thanks for the help. | Tony (4941) | ||
| 1329796 | 2013-02-25 08:53:00 | D'oh! You can do it by Pythagoras: 2802 - 652=74175. SQRT[74175]=272.3508 Q.E.D. | Tony (4941) | ||
| 1 | |||||