Maths In Focus Prelim Ex 6.14, Q8 (1 Viewer)

fifibum · Jun 6, 2016

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!

si2136 · Jun 6, 2016

fifibum said:
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?

eyeseeyou · Jun 6, 2016

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

Dragon9000 · Jun 7, 2016

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

I'll try and solve it when I get back

Green Yoda · Jun 7, 2016

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

Dragon9000 · Jun 7, 2016

Rathin said:
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.

davidgoes4wce · Jun 7, 2016

fifibum said:
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?

Dragon9000 · Jun 7, 2016

davidgoes4wce said:
Are you referring to the old version of the book or the new version?

The Question I posted was from 2nd Edition

davidgoes4wce · Jun 7, 2016

Here is my solution

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

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

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

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

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

$Factorising the h^2 out$

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

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

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

$\therefore h= 83.91 metres$

$Rounded to the nearest metre = 84 metres$

Nicfryk · Mar 3, 2018

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?

Maths In Focus Prelim Ex 6.14, Q8 (1 Viewer)

fifibum

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si2136

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eyeseeyou

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Dragon9000

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Green Yoda

Hi Φ

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davidgoes4wce

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