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thread: Help with Maths question

  1. #1
    Registered User

    Aug 2007
    3,526

    Help with Maths question

    Just wondering if there are any mathameticians out there that can help me with a question!!!

  2. #2
    Registered User

    May 2004
    Shepparton
    4,871

    Might help if you ask the question

  3. #3
    Registered User

    Aug 2007
    3,526

    LOL didnt want to scare anyone away cause it might actually turn into a few questions

    1st question can someone please explain to me sin cos and tan - now i know how to find them but what i don't know is when to use each one - does that make any sense?

  4. #4
    Registered User

    Dec 2005
    6,706

    Maths teacher at your service

    When it comes to trig - SOHCAHTOA is your friend. I have my students write it down every time they see a trig question... What it helps you to remember is that Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse and Tangent is Opposite over Adjacent.

    Does that answer your question?

    I also find that it helps to figure out which one to use if you label the side of the triangle you are using with the letters O, A or H and that helps you see which one you need to use.

    It's really hard doing this without diagrams! If anything's fuzzy, let me know and I'll do my best to clarify for you.

    BW

  5. #5
    Registered User

    Sep 2008
    SE suburbs of Melbourne
    197

    AAAHHHHHH!!

    SOHCAHTOA.....*memories of high school flood her head as she runs screaming out of the room*

  6. #6
    Registered User

    Aug 2007
    3,526

    i understand SOHCAHTOA - but i dont understand when i got to use each one - for example - if a 6m ladder leans against a wall such that it makes an angle of 60degress - how far is the bottom of the ladder away from the wall?

    i have the answer its an example question but i dont know how they come about choosing which one to choose? does that make sense?

  7. #7
    Registered User

    Aug 2007
    3,526

    LOL thanks for your help BabyQ

  8. #8
    Registered User

    Dec 2005
    6,706

    Ok, going to assume that the angle of 60 degrees is between the wall and the ladder, yes?

    First thing, draw a picture and mark in the information you have. The 6m ladder becomes the hypotenuse of the triangle because it's opposite the right angle. The angle of sixty degrees is important - it's the angle you use that decides which side is the opposite and which side is the adjacent. It's the adjacent if it helps form the angle, and it's the opposite if you can draw an arrow across the triangle to point at the side.

    If the angle was between the foot of the ladder and the ground, this next bit is going to be wrong!

    If we call the distance between the foot of the ladder and wall d... then we find that we have a hypotenuse and an unknown opposite side. Opposite and hypotenuse, that tells us it's sin that we use...

    so we write: (and please excuse the lack of symbols!)

    sin 60 = d/6 (that is, opposite side over adjacent side)

    Next step is to simply rearrange the equation to get d by itself... you undo dividing by 6 by multiplying both sides by 6 and end up with:

    d = 6 * sin 60

    and off you go.

    If the angle was between the foot of the ladder and the floor, then the unknown side (d) is actually the adjacent and you'd go with cos...

    cos 60 = d/6, etc.

    Hope that helps clarify things a little more.

    BW

  9. #9
    Registered User

    Aug 2007
    3,526

    okay in that question it was actually cos u had to find, the angle was down the bottom opposite the right angle - but i kinda understand what u said!
    ummm is does the hyp, opp, and adj pair up with either sin, cos or tan?

  10. #10
    Registered User

    Dec 2005
    6,706

    Yeah, once you've figured out which sides of the triangle you are dealing with, it dictates which of the trig ratios you are using. It all comes back to SOHCAHTOA.

    If you have the opposite and adjacent you will always use tan.
    If you have the adjacent and hypotenuse you will always use cos.
    If you have the opposite and hypotenuse you will always use sin.

    That's why it's important to label the sides of the triangle that are important for the question (labelling that third side is a bit redundant, you only want to look for two of O, A and H).

    Find the angle (not the right angle) you are using and figure out which sides are which, set things up properly and you'll never go wrong.

    Remember that SOHCAHTOA not only tells you which sides go with which trig ratio, but which one's on top of the fraction as well.

    Sam's upset, gotta go... hope this makes sense

    BW

  11. #11
    Registered User

    Oct 2007
    Perth,WA
    2,942

    HA...I was SO SO SO SO SO carp at maths in school, but lemme tell ya....SohCahToa is the only thing I remember! I remember the word, but not how to use it though....not like I've ever needed to! Good luck...My head is about to explode just after reading this thread

  12. #12
    Registered User

    Aug 2007
    3,526

    Thankyou -
    Another question for when u get back - what if u only know one side? in this example 1 side was labelled "wall" the other 6m - the other ?m - so only 1 side is known?

    also can i hit u up with how do u find the perpendicular height of a triangle?

  13. #13
    Registered User

    Aug 2007
    3,526

    HA...I was SO SO SO SO SO carp at maths in school, but lemme tell ya....SohCahToa is the only thing I remember! I remember the word, but not how to use it though....not like I've ever needed to!
    LOL the only thing i remember from school is BOMDAS!

  14. #14
    Registered User

    Dec 2005
    6,706

    May do this in stages, depending on how Sam goes in front of the TV.

    Question 1 - I assume you know an angle in that triangle somewhere? I'd assume you're being asked to find the side with the ?m label, but you'd need an angle apart from the right angle somewhere.

    Question 2 - Perpendicular height of a triangle... Is it an equilateral (all sides equal) triangle or an isosceles (two sides equal) triangle? The explanation for how to do those two varies from the explanation for a scalene (no sides equal) triangle and I don't want to confuse you with excess info that you don't need!

    BW

  15. #15
    Registered User

    Aug 2007
    3,526

    Question 1 - the angle is 60

    Question 2 - no sides equal - the sides are 10cm,8cm and 6cm

  16. #16
    Registered User

    Dec 2005
    6,706

    Question 1 - you're good to go. It will be pretty similar to the ladder question we explained earlier, just depends on where the 60 degrees is. Find the side with the question mark.

    Question 2 - still has me a little stumped. With sides of 6, 8 and 10 it's a right angles triangle - there's a right angle between the 6 and the 8 and the 10 is the hypotenuse (have you done much with pythagoras' theorem? 6,8,10 is a pythagorean triad which means it has whole number sides and is a perfect right angle, 3,4,5 is another)... Can you give me the exact wording of the question? And a description of the diagram if there is one?

    Perpendicular height could be either 6 or 8... or if it's the perpendicular height that meets the side of 10 at the base then it's a complicated process of using trig to find the size of one of the angles and then using that angle to determine the height of the triangle. Don't want to go into that if it's one of the more obvious solutions!

    BW

  17. #17
    Registered User

    Aug 2007
    3,526

    okay the point at the top is B - (right angle) than bottom left (looking at it) is A and bottom right is C
    B-A is 6cm
    B-C is 8cm
    A-C is 10cm

    Question is find the area of the triangles using the perpendicular height formula

    another Question is find the area of the triangle ABC - Now to find the area of a triangle isnt it A=1/2 X BASE X PERPENDICULAR HEIGHT?

  18. #18
    Registered User

    Dec 2005
    6,706

    Is there a perpendicular height formula that you have been given? I'm drawing a complete mental blank at the moment and the only information I can find refers to determining the perpendicular height from the area of the triangle and the base length... and if you can't get the area by other means to use Hero's (or Heron's) formula. The mathematical symbols for that one are going to be pretty tough to duplicate here, but you should be able to look it up.

    In any case, the area of the triangle is 24 sq. cm and the perpendicular height is 4.8cm.

    I have to go give Sam a bottle now - I'll check back later.

    BW

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