1. ## Trigonometric ratio problem

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

2. ## Re: 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

Thx a lot.

4. ## Re: Trigonometric ratio problem

Originally Posted by pickslides
In the first part $\displaystyle CD = 250\cos 75$ <--- shouldn't that read 250 * cot(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
I really have some difficulties to understand your solution ....

5. ## Re: Trigonometric ratio problem

so what is the correct answer?? Plz help.

6. ## Re: Trigonometric ratio problem

Originally Posted by william24
so what is the correct answer?? Plz help.
Pickslides got $\displaystyle |\overline{CD}| \approx 64.705\ m$

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

So the values of the speed differ first at the 3rd decimal. Not much harm done.

7. ## Re: Trigonometric ratio problem

I wanna know the steps and how the steps made, My maths is really poor
Thx for ur helping, earboth

8. ## Re: 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.

9. ## Re: Trigonometric ratio problem

Thx. It's clear

10. ## Re: Trigonometric ratio problem

Now $\displaystyle AD=\sqrt{AD^2+BD^2}$
Of course, that should be $\displaystyle AD=\sqrt{AB^2+BD^2}$.