Uploaded it in a PDF, see attachment.

A number of Questions I dont know how to do; hope of getting help to finish this as fast as possible.

Exam & Clumunating Activity Stress in 6 Classes (Angry)

I dont want to fail my math class (Worried)

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- Jan 11th 2009, 09:37 PMSerialkisserApplications of Trigonometry
Uploaded it in a PDF, see attachment.

A number of Questions I dont know how to do; hope of getting help to finish this as fast as possible.

Exam & Clumunating Activity Stress in 6 Classes (Angry)

I dont want to fail my math class (Worried) - Jan 11th 2009, 10:30 PMProve It
- Jan 15th 2009, 06:41 PMSerialkisser
Still having trouble, all this is familiar but it's like something is blocking my brain (Wondering) Having problems with all questions, for the first 3 questions I forgot how to solve for the values if I have a number inside the triangle. I remember a bit of doing something with the 2nd function on the calculator; but I dont even have a scientific calculatoron on me (Rofl)

I am a dead loss !

Now I realize how important it is to listen to the teacher, this teacherless online course kills my brain cells. - Jan 15th 2009, 11:46 PMGrandadSine and Cosine Rule
Hello Serialkisser

In any triangle ABC, side a is opposite angle A, etc, and we have:

Sine Rule: $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

In question 1: you're not interested in a or A, so plug the values of B, C and c into the last two parts of the formula and solve for b.

Cosine Rule: There are three. Here's the first: $\displaystyle a^2 = b^2+c^2 - 2bc \cos A$

The other two just move all the letters around one place: $\displaystyle a \rightarrow b, b \rightarrow c, c\rightarrow a$, etc. So you get:

$\displaystyle b^2 = \dots$ and $\displaystyle c^2 = \dots$

In question 2, you have to make up a similar rule using letters p, q and r. Then plug the values of q, r and P in, and solve for p.

In question 3, write down the Sine Rule using the letters x, y and z, and then use x = 34, X = 110 degrees, z = 15, and solve for Z.

That should get you started.

Grandad

- Jan 16th 2009, 06:45 PMSerialkisser
hmmm I dont undestand (Worried)

Could you solve number 1, and looking at that I will be able to solve number 2 and 3. Just need an example that shown every step :-) - Jan 16th 2009, 07:35 PMProve It
Do you understand the terminology?

If a side is labelled a (say), then the angle opposite the side a is given the symbol A. Similarly side b has an opposite angle labelled B and side c has an opposite angle labelled C.

Can you label the sides and angles?

Can you identify what it is you are trying to find (which side or angle)?

Can you see that it's opposite side or angle is defined.

Can you see that there is another side with its opposite angle defined?

Plug these things into the sine rule and you should be able to solve for the unknown.

Have a go. You have to try these things yourself and make your own mistakes. We can show you where you've made your mistakes and help to rectify them. We're here to help guide you in the right direction, not to do your homework. - Jan 16th 2009, 09:55 PMGrandadTrigonometry
Hello Serialkisser

Here are some worked examples that might help to give you some ideas:

http://www.mathcentre.ac.uk/resource...idkits/4_6.pdf

Solve triangle - SSS

And, if you have Powerpoint, this one is great: http://www.btinternet.com/~mathsanswers/Sine_Rule.ppt

Grandad

- Jan 18th 2009, 06:29 PMTheMasterMind
I'll do the first couple, after that follow what Prove It provided for the others. As he/she said, use the sine law for the first.

To give you more of an idea for when they're used;

1. two angles and any side (sine law)

2. two sides and an angle opposite one of them

3. two sides and a contained angle(cosine law)

4. three sides(cosine)

Alright so for 1/ you have two angles and a side, therefore you must use the sine law.

$\displaystyle \frac{b}{SinB}$ $\displaystyle =\frac{c}{SinC}$

$\displaystyle \frac{b}{Sin72}=\frac{18}{Sin53}$

$\displaystyle \frac{18sin72}{sin53}=21.4cm$

Simply look at what you are given, and make the necessary substitutions into the sine and cosine laws.

For 2/

$\displaystyle p^2=q^2+r^2-2qrCosP$

$\displaystyle

p^2=23^2+12^2-2(23)(12)Cos87$

$\displaystyle

p^2=529+144-552Cos87$

$\displaystyle p^2=673-552Cos87$

$\displaystyle \sqrt{p^2=644.1}$

$\displaystyle p=25.3m$

Follow the same steps and laws for the others - Jan 19th 2009, 09:41 PMSerialkisser
How can I find the second angle in number 3 ?

To find y,I just follow the same steps as number 2.

And I also need help with number 6 (Thinking) - Jan 19th 2009, 10:13 PMGrandadTrigonometry
Hello SerialkisserFind $\displaystyle \theta$ using the Sine Rule, only it's easier if you turn it upside down:

$\displaystyle \frac{\sin\theta}{15}=\frac{\sin 110}{34}$

$\displaystyle \Rightarrow \sin\theta = \frac{15\sin 110}{34}$

When you've got $\displaystyle \theta$, find angle Y using angle sum of triangle = $\displaystyle 180^o$, then use the Cosine Rule:

$\displaystyle y^2 = 15^2 + 34^2 - 2.15.34.\cos Y$

Quote:

And I also need help with number 6 (Thinking)

Draw the diagram as a triangle ABC, marking angle A as the position of the second post, with B and C as the first and third posts. Can you see that angle A = $\displaystyle 120^o$, b = 170 and c = 250? Plug these numbers into the formula, and you've got it.

Grandad