Basic Trigonometry Problem

Beren · 23 May 2007 at 14:51

Ok... assuming it ISNT a right angle triangle and the data you have is-
28 degrees
side of length 4, side of length 7

Then the other two angles would be 124 and 28

and your other side would be 3.9

The question in that paper is impossible I'm afraid!

If you don't believe me, try to draw it to scale... I'm not suprised you were struggling with it!

georges · 23 May 2007 at 14:51

With what I've learnt I can't see it working..

What level is it?

a^2 = 4^2 + 7^2 -2x4x7 cos(28)

would suggest it was 3.94

But yeah, as above, that isn't a triangle, or it isn't right angled at least.

Mr.Clark · 23 May 2007 at 14:52

tntcoder said:
It is a right angle triangle though? Heres a direct copy of the exam question

Are you saying those values dont work at all?

You don't even need Trig for that.

Pythag gets you a (a^2 = 7^2 - 4^2) and (180-90-28) gets you the angle.

Simple.

The little square in the bottom left shows that it is a right angled triangle.

georges · 23 May 2007 at 14:53

Except they're wrong?

if it did work, 7sin28 would = 4, except it doesn't..

KaHn · 23 May 2007 at 14:55

georges said:
Except they're wrong?

if it did work, 7sin28 would = 4, except it doesn't..

Thats the only part im gettin annoyed at.

KaHn

I've got it, this is just a mistake with the right angle symbol, thus normal trig rules dont work and Pyth doesn't either.

Back to trig idents

georges · 23 May 2007 at 14:57

The question, not me?

Mr.Clark · 23 May 2007 at 14:58

It should be 5, not 7... ?

gurusan · 23 May 2007 at 15:00

It's true! The question is impossible!

B = 180 - (90 + 28)
B = 62

cos(62) = a/7
7cos(62) = a

a = 3.28630094

Check with Pythagorean Theorem:

a^2+16=49
a^2 = 33
sqrt(33) = 5.744562647

NO WORKY!

Ladforce · 23 May 2007 at 15:01

Yes there is an error in the question as that problem, with the constraints set does not work.

gurusan · 23 May 2007 at 15:01

I would recommend showing them the exam question was BS.

tntcoder · 23 May 2007 at 15:02

Hmm thanks a lot guys, didnt know there was an error in the question :rolleyes:

, I understand the theory anyway, just didnt get why the trig wasnt working.

Beren · 23 May 2007 at 15:03

Do you have to hand this in or is it revision?

If you have to hand it in then I would use mr Clarks quotation! Otherwise ignore and move on.

georges · 23 May 2007 at 15:05

Not got the markscheme to hand, or the exam chief's report?

tntcoder · 23 May 2007 at 15:05

No its just last years past paper, theres only 1 question on basic trig, so i wont worry too much about it and will get on to the harder matrix and vector work.

KaHn · 23 May 2007 at 15:09

The question works fine if you assume that the other angle isnt 90.

Just work out as you would with the laws ive stated, ah nuts to it ill work it out and post my working.

KaHn

Mr.Clark · 23 May 2007 at 15:11

KaHn said:
The question works fine if you assume that the other angle isnt 90.

That's the trouble though.

The question also works fine if you assume that the angle isn't 28 degrees. Or the right angle isn't. Or the 4 is wrong. Or the 7 is wrong.

How does anyone know which assumption to make?

To be brutally honest, the question should have been voided. To be fair, they should have accepted any answer where one of our assumptions was stated and the answer worked out from there...

KaHn · 23 May 2007 at 15:19

KaHn

KaHn · 23 May 2007 at 15:21

Mr.Clark said:
Snip

For Degree I would expect it to be a bit harder than pythag and basic tri tho.

My working is if you dont use a 90deg triangle.

Aslong as you state your assumptions he will have to give you marks for working.

KaHn

BillytheImpaler · 23 May 2007 at 15:28

As everybody else has said the question is flawed as the triangle it proposes is impossible. Here's another graphical non-solution:

Darg · 23 May 2007 at 15:33

Make a complaint and have the entire year from before come in to resit their exams.. they'll love you

Basic Trigonometry Problem

Man of Honour