1. ## Geometric trigonometry

Can someone please help me with the following problem? I have absolutely no clue as to how to do this!

2. ## Trigonometry

Hello Nancy

In question 1, D is missing, but I assume it's where the line from C meets AB.

So, look at $\triangle BCD$:

$DC = BC\cos x, \quad BD = BC\sin x$

So $BC(BC+BD)=2DC^2$

$\Rightarrow BC(BC+BC\sin x)= 2BC^2\cos^2x$

$\Rightarrow BC^2(1+\sin x) = BC^2 \times 2\cos^2 x$

$\Rightarrow 2\cos^2 x = 1 + \sin x$

Now use $\cos^2x = 1 - \sin^2x$:

$2(1-\sin^2x) = 1 + \sin x$

$\Rightarrow 2\sin^2x + \sin x -1 =0$

$\Rightarrow (2\sin x -1)(\sin x +1) = 0$

$\Rightarrow \sin x = \tfrac12$, since $\sin x = -1$ is impossible in this case.

$\Rightarrow x = 30^o$

I think you may have written down some of the other questions wrong as well. But here are a few pointers.

2 Note that $\angle CAD = x$, and then express all the lengths in terms of $AD$. So:

$AB = AD\csc x, \quad AC = AD\sec x, \quad BC = BD+DC = ...$?

Substitute all these values into $3AB+2AC=2AD + BC$, and divide through by $AD$.

I think you have written 2(b) wrong. It doesn't work out as you have written it.

3 Express $BD$ and $BC$ in terms of $AB$. Again, I think you haven't written the result correctly. I think it should be $2\cos^4x-3\cos^2x + 1 = 0$

The result then factorises: $(2\cos^2x-1)(\cos^2x-1) = 0$

4 Should there be some more information in the diagram? Have you written the result correctly?

3. Sir,
Thank u sooo much for ur response! It was a great help!

Now that you have pointed this out, it turns out that the original diagrams did have some mistakes:

1) In question 2 (b), I don't think it is very clear on the pdf version but the sign between 3, cosx and sinx is a minus sign.
So, the expression is tanx = 3 - cosx - sinx
sinx + cosx - 2

2) The homework sheet says 2cos^4 x - 3cosx+1 = 0
But now that I look at it (thank u for ur help again) it is probably meant to be cos^2 x. Our teacher probably missed it (He has a habbit of doing that).

3) And yes, the C was probably meant to be a D and vice-versa. But everything else seems ok (but again, there could have been a miss print)

Thank u again!

Nancy