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Some more study material

  1. Oct 27, 2013 #1
    Hello to all i am new to this thread and i have a precalculus exam tomorrow. I was going to post 4 or 5 example
    problems that i have and wanted to know if someone could explain to me how to get the answer step by step.
    I need to use these explanations as a help guide and as long as its step by step i can understand them
    better. thank you very much in advance for those who can help me.

    i need some help with these type of problems it deals with angles im lost please could someone explain with as much detail as possible.

    1.) in traingle ABC , B =65(degrees) C=45(degrees) and c = 12 what is the measure of side b?(two decimal places)

    2.) in traingle ABC , B =90(degrees) C=30(degrees) and c = 15 what is the exact measure of side b?

    3.) in traingle ABC , A =45(degrees) B=60(degrees) and a = 9 .side b to the nearest hundredth?


    4.) in traingle ABC , a =10 b=20 and c = 13 what is the measure of side B to the nearest 10th degree?
     
  2. jcsd
  3. Oct 27, 2013 #2

    Mentallic

    User Avatar
    Homework Helper

    You need to start getting help for your exam earlier than the day before! On this forum we aren't allowed to just provide solutions.. What we can do however is help you along wherever you get stuck.


    For all of these problems, you want to use the sine rule:

    [tex]\frac{a}{\sin{A}}=\frac{b}{\sin{B}}[/tex]

    What this formula is saying is that if you have the 3 of the 4 properties of a triangle:
    1. length of one side (we'll call it a, but it doesn't have to be the 'a' in your triangles)
    2. the opposite angle to a which we denote A
    3. another side length b
    4. the opposite angle B

    then you can use the formula to find the answer to the last property of the triangle that you need to find. For your first question

    1.) in traingle ABC , B =65(degrees) C=45(degrees) and c = 12 what is the measure of side b?(two decimal places)

    we will use the sine rule

    [tex]\frac{b}{\sin{B}} = \frac{c}{\sin{C}}[/tex]

    plugging in our values B, C and c, we get

    [tex]\frac{b}{\sin{65^o}}=\frac{12}{\sin{45^o}}[/tex]

    and now we need to solve for b. Can you finish this off? It's basically the same deal for the other 2 questions as well.

    You need to use the cosine rule for this one.

    [tex]a^2=b^2+c^2-2bc\cdot\cos{A}[/tex]

    but we need to find the value of an angle, so we need to make [itex]\cos{A}[/itex] the subject of this formula:

    [tex]a^2-b^2-c^2=-2bc\cdot\cos{A}[/tex]

    [tex]\cos{A}=\frac{a^2-b^2-c^2}{-2bc}[/tex]

    [tex]\cos{A}=\frac{b^2+c^2-a^2}{2bc}[/tex]

    now you just need to plug in the values, but notice that this formula is finding [itex]A[/itex] while you are looking for the angle B, so you could swap the a's for the b's (because the letters in these formulae are arbitrary, all you need to keep consistent is the lowercase letter side length with its uppercase letter opposite angle). So

    [tex]\cos{B} = \frac{a^2+c^2-b^2}{2ac}[/tex]
     
  4. Oct 28, 2013 #3

    Mark44

    Staff: Mentor

    mathopressor,
    Read your private mail. You have already received one warning for posting a thread without showing any work. This is a violation of Physics Forums rules, the rules you agreed to follow when you opened your account.

    I am closing this thread.
     
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