Trig. ratios and trig expressions

• Jun 21st 2010, 04:40 PM
sblocker
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
• Jun 22nd 2010, 09:21 AM
Hello sblocker

Welcome to Math Help Forum!
Quote:

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 $60^o$, not $6^o$. Look at the diagram (click to enlarge):
Attachment 17961
Can you see that you'll use the angle of $30^o$ to find the ratios for $210^o$?

Quote:

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 $30^o$ and $60^o$. (You should know what these are.)

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

Quote:

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.

• Jun 22nd 2010, 10:58 AM
Hello again sblocker
Quote:

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:
Attachment 17963
(a) The triangle is right-angled at L, and, since it is isosceles (LY = LF), $\angle F = 45^o$.

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

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

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