Hey, I have a whole sheet of trigonometry questions to do and I've drawn just 4 of them in paint. Please take me through them as I haven't a clue. Many thanks!

http://img393.imageshack.us/img393/3381/maths0sv.jpg

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- Jan 13th 2006, 08:41 AMtristan1Trigonometry
Hey, I have a whole sheet of trigonometry questions to do and I've drawn just 4 of them in paint. Please take me through them as I haven't a clue. Many thanks!

http://img393.imageshack.us/img393/3381/maths0sv.jpg - Jan 13th 2006, 08:45 AMTD!
In a right triangle, the sine of one of the (non-right) angles is equal to the opposite side divided by the hypotenuse while the cosine of that angle is the adjacent side divided by the hypotenuse.

For problem 1, this gives that**cos(27°) = x/5.89**which is an equation you can solve for x. The other problems are similar! - Jan 13th 2006, 08:50 AMtristan1
So my answer for 1 would be x= 5.2480 (4dp)

Am I right?

Can you give me the equation used, as you did for cos(27°) = x/5.89, for the others please? - Jan 13th 2006, 08:54 AMTD!Quote:

Originally Posted by**tristan1**

Quote:

Originally Posted by**tristan1**

Try to set-up such an equation youself. I'll list the possible formula's you can use:

$\displaystyle \begin{gathered}

\sin \alpha = \frac{{{\text{opposite}}}}

{{{\text{hypotenuse}}}} \hfill \\

\cos \alpha = \frac{{{\text{adjacent}}}}

{{{\text{hypotenuse}}}} \hfill \\

\tan \alpha = \frac{{{\text{opposite}}}}

{{{\text{adjacent}}}} \hfill \\

\end{gathered} $

Now: first identify which of the three is given (the opposite, adjacent or hypotenuse) and which is asked.

Then choose the formula which has both these sides and try to set up the equation. - Jan 13th 2006, 09:05 AMtristan1
OK then, can you explain why I used COS for question one.

I really dont get how you determin which equation you use. - Jan 13th 2006, 09:10 AMTD!
Of course, as I said: you have to determine which sides are involved in the problem.

Do you understand the terms adjacent, opposite and hypotenuse?

Here's a clear image, be careful because it's important which angle is marked.

http://www.gcse.com/maths/images/hypotenuse.gif

Now, for the marked angle at the right, the bottom side is 'adjacent' (it touches the angle but isn't the longest side), the left side is 'opposite' (it doesn't touch the angle) and the longest side is called the hypotenuse.

In the first problem, the marked angle is at the top left. The side we're looking for is the top side, which is the adjacent side for this angle. The hypotenuse is given, so we're looking for the formula which involves the**hypotenuse**and the**adjacent**side, that's the on with the cosine.

Do you understand? - Jan 13th 2006, 09:31 AMtristan1
OK, I kinda understand. For 2 (and 3), I need to find the hypotonuse.

So I'm looking for one of those 3 equations to contain the ajacent and hypotenuse. Therefore I'd use COS?

For 2, I'd type in COS 55 in the calculator then times the answer by 45?

For 3, I'd type COS 71 in the calculator the tiems by 163. Are these correct? - Jan 13th 2006, 09:34 AMTD!
Take them one at a time and look carefully.

For two, the marked angle is at the top left. The hypotenuse is indeed the side we're looking for, but which one is given (from the point of view of the angle): the adjacent or opposite side? - Jan 13th 2006, 09:36 AMtristan1
Opposite?

- Jan 13th 2006, 09:37 AMTD!
Indeed! The side which isn't linked to the angle, that is the opposite side.

So what formula will you be using? You need one that involed the hypotenuse and the opposite side. - Jan 13th 2006, 09:39 AMtristan1
SIN?

So how do I approach this in a calculator? - Jan 13th 2006, 10:04 AMTD!
Indeed sin. Now write out the formula I gave for the sine, replacing all expressions you know with their values and call the unknown side

**x**.

Then solve for x, the expression you now have can be entered in a calculator. - Jan 13th 2006, 10:10 AMtristan1
SIN 55 =0.8191

0.8191 x 45 = 36.8618

Right? - Jan 13th 2006, 10:13 AMTD!
You should first try to set-up the equation correctly, starting from the formula.

Write down the formula of the sine and fill in eveything you know, i.e. replace "opposite", "hypotenuse" and "alpha" with their values (or x, if it's the unknown).

$\displaystyle \sin \alpha = \frac{{{\text{opposite}}}}

{{{\text{hypotenuse}}}} \Rightarrow \sin 55^\circ = \frac{{45}}

{x}$

You then solve for x

$\displaystyle \sin 55^\circ = \frac{{45}}

{x} \Leftrightarrow x\sin 55^\circ = 45 \Leftrightarrow x = \frac{{45}}

{{\sin 55^\circ }} \approx 54.93$