Results 1 to 10 of 10

Math Help - lenghts of sides triangle

  1. #1
    Senior Member
    Joined
    Sep 2009
    Posts
    300

    lenghts of sides triangle

    Hello,
    Can you tell me how to find the sides of he triangles in Q1 and Q2 in the attached drawing
    they are right angled triangles
    in question 2 i found that a=1.9 and b=1.3 using Soh

    but i want to know If i can use a^2 = b^2 + c^2 - 2bc(CosA)

    I want to be able to use other forumlas than a^2 + b^2=c^2

    Please help thanks
    Attached Thumbnails Attached Thumbnails lenghts of sides triangle-triangleprob.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Oct 2009
    Posts
    459
    I don't think any values can be found for example 2.

    The first example has a right angle and you don't want to use more complicated methods than is necessary so the Pythagoras theorem will do or the basic trigonometrical ratios.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    There is answers! I dont know how to get them someone help!!!!!!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,061
    Thanks
    900
    Quote Originally Posted by wolfhound View Post
    Hello,
    Can you tell me how to find the sides of he triangles in Q1 and Q2 in the attached drawing
    they are right angled triangles
    in question 2 i found that a=1.9 and b=1.3 using Soh

    but i want to know If i can use a^2 = b^2 + c^2 - 2bc(CosA)

    I want to be able to use other forumlas than a^2 + b^2=c^2

    Please help thanks
    as stated previously by the Stuck Man ... question 2 does not have enough information.

    you need one more side or one more angle. there are an infinite number of right triangles with hypotenuse length 2.3
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    Sorry,
    The angle B in question 2 is 34 degrees
    Please tell me how to calculate the answers for the two questions,I have an important exam next week, can it be done with the other formula in my original post?I can seem to do them with that?is that formula only for triangles that are not 90*
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by wolfhound View Post
    Sorry,
    The angle B in question 2 is 34 degrees
    Please tell me how to calculate the answers for the two questions,I have an important exam next week, can it be done with the other formula in my original post?I can seem to do them with that?is that formula only for triangles that are not 90*
    Q1. It's an isosceles triangle therefore c = 1.

    Q2. You should know this:

    \sin 34^0 = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{b}{2.3} \Rightarrow ....

    \cos 34^0 = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{a}{2.3} \Rightarrow ....
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    Quote Originally Posted by mr fantastic View Post
    Q1. It's an isosceles triangle therefore c = 1.

    Q2. You should know this:

    \sin 34^0 = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{b}{2.3} \Rightarrow ....

    \cos 34^0 = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{a}{2.3} \Rightarrow ....
    Yes thats how I got the sides on question 2.

    But I want to know how to work out the other side is 1(in question 1). In the exam I cant just write 1,I need to prove it

    When is this formula used a^2 = b^2 + c^2 - 2bc(CosA)
    and why cant I find the other side using this?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by wolfhound View Post
    Yes thats how I got the sides on question 2.

    But I want to know how to work out the other side is 1(in question 1). In the exam I cant just write 1,I need to prove it

    When is this formula used a^2 = b^2 + c^2 - 2bc(CosA)
    and why cant I find the other side using this?
    Did I not I say
    It's an isosceles triangle
    which, I would have thought, provides the necessary justification.

    Alternatively, use Pythagoras' Theorem. Alternatively, use \sin 45^0 = \frac{\text{opposite side}}{\text{hypotenuse}} (which you should have realised from what I posted about question 2).

    Why are you insisting on wanting to use the cosine rule? For a right angle triangle A = 90 degrees and it's just Pythagoras' Theorem. The cosne rule and the sine rule are used when you have a triangle that is NOT a right-triangle. You should know that.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Sep 2009
    Posts
    300
    Quote Originally Posted by mr fantastic View Post
    Did I not I say

    which, I would have thought, provides the necessary justification.

    Alternatively, use Pythagoras' Theorem. Alternatively, use \sin 45^0 = \frac{\text{opposite side}}{\text{hypotenuse}} (which you should have realised from what I posted about question 2).

    Why are you insisting on wanting to use the cosine rule? For a right angle triangle A = 90 degrees and it's just Pythagoras' Theorem. The cosne rule and the sine rule are used when you have a triangle that is NOT a right-triangle. You should know that.
    I am curious why I could not get a correct answer using the cosine rule? I thought it would work on right angled triangles also? NO?
    I want to get in practice with the different formulas that is all
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Dec 2009
    Posts
    1
    Here are links to tutorials for basic "right angle" trig, with a very DIFFERENT approach than the usuals "soh cah toa" mneumonic. The first shows you how to make a "Trig Tool", which elimininates memorizing a lot of formulas, the next two show you how to USE it.

    How to Memorize the Trig Functions (Without Losing Your Mind!) | eHow.com
    How to Pass That Trigonometry Exam (Without Losing Your MIND!) Part I - Unknown Sides | eHow.com
    How to Pass That Trigonometry Exam (Without Losing Your MIND!) Part II - Unknown Angles | eHow.com
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relation between cosines and lenghts of triangle
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: July 19th 2011, 11:12 AM
  2. Replies: 2
    Last Post: June 27th 2010, 11:25 AM
  3. Replies: 0
    Last Post: April 9th 2010, 07:53 PM
  4. Replies: 7
    Last Post: July 19th 2008, 07:53 AM
  5. three sides of a triangle
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 1st 2008, 08:22 AM

Search Tags


/mathhelpforum @mathhelpforum