# Thread: Trig. ratios and trig expressions

1. ## Trig. ratios and trig expressions

hi so i need help for these Two questions:

1. a) To find trigonometric ratios for 240 degrees using a unit circle, a Reference angle of 6 degrees is used. what reference angle should you use to find the trigonometric ratios for 210 degrees?

b) Use the unit circle to find exact values of the three primary trig. ratios for 210 and 240 degrees

2. A fishing boat is 15km south of a lighthouse. A yacht is 15km west of the same lighthouse.

a) Use trigonometry to find an exact expression for the distance between the two boats.

Any help would be appreciated

2. Hello sblocker

Welcome to Math Help Forum!
Originally Posted by sblocker
hi so i need help for these Two questions:

1. a) To find trigonometric ratios for 240 degrees using a unit circle, a Reference angle of 6 degrees is used. what reference angle should you use to find the trigonometric ratios for 210 degrees?
I think you mean $\displaystyle 60^o$, not $\displaystyle 6^o$. Look at the diagram (click to enlarge):

Can you see that you'll use the angle of $\displaystyle 30^o$ to find the ratios for $\displaystyle 210^o$?

b) Use the unit circle to find exact values of the three primary trig. ratios for 210 and 240 degrees
Write down the sine, cosine and tangent of $\displaystyle 30^o$ and $\displaystyle 60^o$. (You should know what these are.)

Can you see from the diagram that to get the sine, cosine and tangent of $\displaystyle 210^o$ and $\displaystyle 240^o$ you'll simply change the sign of all these values from + to - ?

2. A fishing boat is 15km south of a lighthouse. A yacht is 15km west of the same lighthouse.

a) Use trigonometry to find an exact expression for the distance between the two boats.

Any help would be appreciated
No time to answer this now. I'll do so later if no-one else answers in the meantime.

3. Hello again sblocker
Originally Posted by sblocker
2. A fishing boat is 15km south of a lighthouse. A yacht is 15km west of the same lighthouse.

a) Use trigonometry to find an exact expression for the distance between the two boats.

Any help would be appreciated
Take a look at the diagram:

(a) The triangle is right-angled at L, and, since it is isosceles (LY = LF), $\displaystyle \angle F = 45^o$.

So:
$\displaystyle \sin 45^o = \dfrac{LY}{FY}$

$\displaystyle \Rightarrow FY=\dfrac{LY}{\sin 45^o}$
$\displaystyle =15\div\dfrac{1}{\sqrt2}$

$\displaystyle =15\sqrt2$
(b) Using Pythagoras' Theorem:
$\displaystyle FY^2 = 15^2+15^2$
$\displaystyle =15^2\times2$
$\displaystyle \Rightarrow FY = 15\sqrt2$