# Trigonometric ratio problem

• Aug 19th 2012, 08:18 PM
william24
Trigonometric ratio problem
I don't know how to do the questions below↓ Can anyone help me to solve this question? Thanks.
Attachment 24552
• Aug 19th 2012, 08:27 PM
pickslides
Re: Trigonometric ratio problem
In the first part $CD = 250\cos 75$

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

Speed = Distance $\div$ Time = $\frac{\sqrt{ 350^2+ (300+250\cos 75)^2} }{12\times 60}$ metres per second
• Aug 19th 2012, 08:41 PM
william24
Re: Trigonometric ratio problem
Thx a lot.
• Aug 19th 2012, 11:13 PM
earboth
Re: Trigonometric ratio problem
Quote:

Originally Posted by pickslides
In the first part $CD = 250\cos 75$ <--- shouldn't that read 250 * cot(75°)

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

Speed = Distance $\div$ Time = $\frac{\sqrt{ 350^2+ (300+250\cos 75)^2} }{12\times 60}$ metres per second

I really have some difficulties to understand your solution ....
• Aug 20th 2012, 02:48 AM
william24
Re: Trigonometric ratio problem
so what is the correct answer?? Plz help.
• Aug 20th 2012, 03:05 AM
earboth
Re: Trigonometric ratio problem
Quote:

Originally Posted by william24
so what is the correct answer?? Plz help.

Pickslides got $|\overline{CD}| \approx 64.705\ m$

I got $|\overline{CD}| \approx 66.99\ m$

So the values of the speed differ first at the 3rd decimal. Not much harm done.
• Aug 20th 2012, 03:12 AM
william24
Re: 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, 08:26 AM
Re: Trigonometric ratio problem
The steps are the following. You need to find Dickie's walking speed.

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

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:

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

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

Now $AD=\sqrt{AD^2+BD^2}=\sqrt{350^2+(300+250\cdot\cot( 75))^2}$,
so the speed is $\sqrt{350^2+(300+250\cdot\cot(75))^2}/12$ 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, 06:41 PM
william24
Re: Trigonometric ratio problem
Thx. It's clear
• Aug 21st 2012, 04:32 AM
Now $AD=\sqrt{AD^2+BD^2}$
Of course, that should be $AD=\sqrt{AB^2+BD^2}$.