I don't know how to do the questions below↓ Can anyone help me to solve this question? Thanks.

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Please help, thx.

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- Aug 19th 2012, 07:18 PMwilliam24Trigonometric ratio problem
I don't know how to do the questions below↓ Can anyone help me to solve this question? Thanks.

Attachment 24552

Please help, thx. - Aug 19th 2012, 07:27 PMpickslidesRe: Trigonometric ratio problem
In the first part $\displaystyle CD = 250\cos 75$

Therefore the distance $\displaystyle AD = \sqrt{ 350^2+ (300+250\cos 75)^2} $

Speed = Distance $\displaystyle \div $ Time = $\displaystyle \frac{\sqrt{ 350^2+ (300+250\cos 75)^2} }{12\times 60}$ metres per second - Aug 19th 2012, 07:41 PMwilliam24Re: Trigonometric ratio problem
Thx a lot.

- Aug 19th 2012, 10:13 PMearbothRe: Trigonometric ratio problem
- Aug 20th 2012, 01:48 AMwilliam24Re: Trigonometric ratio problem
so what is the correct answer?? Plz help.

- Aug 20th 2012, 02:05 AMearbothRe: Trigonometric ratio problem
- Aug 20th 2012, 02:12 AMwilliam24Re: Trigonometric ratio problem
I wanna know the steps and how the steps made, My maths is really poor :(

Thx for ur helping, earboth :) - Aug 20th 2012, 07:26 AMtomasklosRe: Trigonometric ratio problem
The steps are the following. You need to find Dickie's walking speed.

$\displaystyle \mbox{walking speed} = \frac{\mbox{distance travelled}}{\mbox{time travelled}}$.

You know "time travelled", so you only need to find "distance travelled."

You can apply Pythagoras' theorem to find the distance from D to A if you know AB and BD.

AB is given (350 meters), and BD = CD + 300 meters, so you need to find CD, which you can derive from the fact that CE as well as the angle at D are given:

$\displaystyle \tan(75) = \frac{EC}{CD}$,

so $\displaystyle CD\cdot\tan(75) = EC$, and $\displaystyle CD=\frac{EC}{\tan(75)}=EC\cdot\cot(75)=250\cdot \cot(75)$.

Now $\displaystyle AD=\sqrt{AD^2+BD^2}=\sqrt{350^2+(300+250\cdot\cot( 75))^2}$,

so the speed is $\displaystyle \sqrt{350^2+(300+250\cdot\cot(75))^2}/12$ meters per minute. (Convert to whatever unit required.)

Try to understand the steps above, and then solve problem (b) on your own. - Aug 20th 2012, 05:41 PMwilliam24Re: Trigonometric ratio problem
Thx. It's clear

- Aug 21st 2012, 03:32 AMtomasklosRe: Trigonometric ratio problem