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I'm stuck on maths :-(

ProfilePosted byOptionsPost Date

Len of the Chilterns

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

Luciacw Report 30 Mar 2006 23:29

lol Fred %3A-%29

Luciacw

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!! %3A-%29

Wulliam

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

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

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 %3A-%29

Luciacw

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.. %3A-%29

Pete

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

Luciacw Report 30 Mar 2006 15:30

Thanks Julie %3A-%29 The problem is if 17(cos45cosx + sin45sinx) = the height of the triangle, How am to find the exact value bearing in mind I'm supposed to be able to work this out in my head without a calculator? :-s This is soo confusing! Thank you very much for your help, %3A-%29

Luciacw

Luciacw Report 30 Mar 2006 15:21

Thanks Julie, I'll go and have a look

Unknown

Unknown Report 30 Mar 2006 15:18

Try this site:- http://mathworld.wolfram(.)com/IsoscelesTriangle.html Julie xxx

Luciacw

Luciacw Report 30 Mar 2006 15:15

oh well back to the drawing board..maths is too hard for me %3A-%28

Luciacw

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