General Chat
Welcome to the Genes Reunited community boards!
- The Genes Reunited community is made up of millions of people with similar interests. Discover your family history and make life long friends along the way.
- You will find a close knit but welcoming group of keen genealogists all prepared to offer advice and help to new members.
- And it's not all serious business. The boards are often a place to relax and be entertained by all kinds of subjects.
- The Genes community will go out of their way to help you, so don’t be shy about asking for help.
Quick Search
Single word search
Icons
- New posts
- No new posts
- Thread closed
- Stickied, new posts
- Stickied, no new posts
I'm stuck on maths :-(
Profile | Posted by | Options | Post Date |
---|---|---|---|
|
Len of the Chilterns | Report | 30 Mar 2006 23:34 |
Lucia Its 66 years since I had that problem. been scratching my head for 5 minutes but the answer won't come. len |
|||
|
Luciacw | Report | 30 Mar 2006 23:29 |
lol Fred |
|||
|
Luciacw | Report | 30 Mar 2006 23:28 |
Hello William,
I don't know why I said the sides were parallel when I meant congruent! lol :-s
My dad managed to solve after about an hour. He said this is the hardest maths problem I've asked him to help me solve lol.
It turns out using a=bh, and some substitution into trigonometric formulae, the area works out as 80.5cm.
sin (45 -x) = cosx/root2 - sinx/root2 = base/17
cos (45-x) = cosx/root2 + sinx/root2= height/17
area = 2.17.17.(cosx/root2 + sinx/root2) (cosx/root2 - sinx/root2)
= (15/17root2 - 8/17root2) (23/17root2)
=161/2 = 80.5 square cm...PHEW!!
|
|||
|
Wulliam | Report | 30 Mar 2006 17:48 |
Hi Lucia, You say: 17(cos45cosx + sin45sinx) = the height of the triangle In that case 17cos(45 - x) = height of triangle This is because of the trig result: cos(A-B) = cosAcosB + sinAsinB I'm not sure of the rest (even though I teach Maths) because you mention that this triangle has parallel sides...! Did you mean something else? Regards, William |
|||
|
Pete | Report | 30 Mar 2006 16:06 |
So half the base can be calculated: Sin (angle) 161/289 = opp/hyp = opp/17 17 x 161/289 = Opp (half the base) So 161/17 cm for half the base Now use pythag to cal the height Once height is done work out area. Pete |
|||
|
Luciacw | Report | 30 Mar 2006 15:53 |
Hi Pete, just checked the book and according to that it is161/289..This seems impossible!
Thank you very much for your help
|
|||
|
Luciacw | Report | 30 Mar 2006 15:40 |
It's difficult without a diagram but..if the triangle is ABC, then,
^ABC = sin(90-2x) = 161/289..the angle is at the top of the diagram.
the side on the left hand side is 17cm and it is parallel to another side, on the right..
|
|||
|
Pete | Report | 30 Mar 2006 15:36 |
Lucia, In triangle ABC which angle do you know? Which two angles are equal. Which side do you have? Which two sides are equal? |
|||
|
Luciacw | Report | 30 Mar 2006 15:30 |
Thanks Julie |
|||
|
Luciacw | Report | 30 Mar 2006 15:21 |
Thanks Julie, I'll go and have a look |
|||
|
Unknown | Report | 30 Mar 2006 15:18 |
Try this site:- http://mathworld.wolfram(.)com/IsoscelesTriangle.html Julie xxx |
|||
|
Luciacw | Report | 30 Mar 2006 15:15 |
oh well back to the drawing board..maths is too hard for me |
|||
|
Luciacw | Report | 30 Mar 2006 15:07 |
I have an isosceles triangle. I know that one of the angles, sin(90 -2x) = 161/289 and that one of the sides is 17cm long. How do I find the area of that triangle. :-s |