# Thread: Maths In Focus Prelim Ex 6.14, Q8

1. ## Maths In Focus Prelim Ex 6.14, Q8

I'm working from the Maths In Focus prelim ext. 1 textbook (Grove)
and am so stuck on ex 6.14, question 8...cosine rule doesn't seem to be
working...help!

2. ## Re: Maths In Focus Prelim Ex 6.14, Q8

Originally Posted by fifibum
I'm working from the Maths In Focus prelim ext. 1 textbook (Grove)
and am so stuck on ex 6.14, question 8...cosine rule doesn't seem to be
working...help!
Have you tried drawing a diagram to begin with?

3. ## Re: Maths In Focus Prelim Ex 6.14, Q8

Post this in the maths forum. Not here (the extension 1 maths thread)

4. ## Re: Maths In Focus Prelim Ex 6.14, Q8

It is an extension 1 question, with 3D trig.

I'll try and solve it when I get back

5. ## Re: Maths In Focus Prelim Ex 6.14, Q8

Tell/Show us the question, we dont all have MIF

6. ## Re: Maths In Focus Prelim Ex 6.14, Q8

Originally Posted by Rathin
Tell/Show us the question, we dont all have MIF
David walks along a straight road. At one point he notices a tower on a bearing of 053° with an angle of elevation of 21°. After walking 230 m, the tower is on a bearing of 342°, with an angle of elevation of 26°. Find the height of the tower correct to the nearest metre.

I'm not OP, but I didn't really know where to start, I drew a shit diagram though.

7. ## Re: Maths In Focus Prelim Ex 6.14, Q8

Originally Posted by fifibum
I'm working from the Maths In Focus prelim ext. 1 textbook (Grove)
and am so stuck on ex 6.14, question 8...cosine rule doesn't seem to be
working...help!
Are you referring to the old version of the book or the new version?

8. ## Re: Maths In Focus Prelim Ex 6.14, Q8

Originally Posted by davidgoes4wce
Are you referring to the old version of the book or the new version?
The Question I posted was from 2nd Edition

9. ## Re: Maths In Focus Prelim Ex 6.14, Q8

Here is my solution

$tan \ 21^\circ=\frac{h}{x} \implies \ x=\frac{h}{tan \ 21^\circ}$

$tan \ 26^\circ=\frac{h}{y} \implies \ y=\frac{h}{tan \ 26^\circ}$

$Using the cosine rule and applying it to the base of Triangle ABC$

$230^2=x^2+y^2-2xy \ cos 71^\circ$

$230^2=(\frac{h}{tan\ 21^\circ})^2+(\frac{h}{tan \ 26^\circ})^2-2(\frac{h}{tan \ 21^\circ}) (\frac{h}{tan \ 26^\circ}) cos \ 71^\circ$

$Factorising the h^2 out$

$230^2=h^2(\frac{1}{tan^{2} \ 21^\circ}+\frac{1}{tan^{2} \ 26^\circ}-\frac{2 \ cos 71^\circ}{tan 21^\circ \ tan 26^\circ})$

$\frac{230^2}{(\frac{1}{tan^{2} \ 21^\circ}+\frac{1}{tan^{2} \ 26^\circ}-\frac{2 \ cos 71^\circ}{tan 21^\circ \ tan 26^\circ})}=h^2$

$\sqrt{\frac{230^2}{(\frac{1}{tan^{2} \ 21^\circ}+\frac{1}{tan^{2} \ 26^\circ}-\frac{2 \ cos 71^\circ}{tan 21^\circ \ tan 26^\circ})}}=h$

$\therefore h= 83.91 metres$

$Rounded to the nearest metre = 84 metres$

10. ## Re: Maths In Focus Prelim Ex 6.14, Q8

Why doesn't it work if I use the Sine Rule to find your x and y above and then substitute those into the tan ratio?

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