Results 1 to 5 of 5

Math Help - help me solve this considered easy maths question

  1. #1
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177

    Post help me solve this considered easy maths question

    the AC is not a straight line by the way. Just afraid my drawings are not clear
    Attached Thumbnails Attached Thumbnails help me solve this considered easy maths question-img031.bmp  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by helloying View Post
    the AC is not a straight line by the way. Just afraid my drawings are not clear
    see the modified diagram below. Note that AB = E + F, where E and F are the bases of the two right-triangles we've constructed. we can find their lengths using trig ratios:

    \sin 40^o = \frac E8

    \cos 25^o = \frac F9

    EDIT: i forgot about finding the angle CAD. Note that you can find the length of BC in the same way we found AB. once that is found, we can use Pythagoras' theorem to find the length AC. Thus we will know the lengths of all the sides of triangle ACD, and we can use the law of cosines to find any angle in that triangle we want.
    Attached Thumbnails Attached Thumbnails help me solve this considered easy maths question-tri.bmp  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member helloying's Avatar
    Joined
    Jul 2008
    Posts
    177
    oh i see thanks for ur help. u are so smart.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by helloying View Post
    oh i see thanks for ur help. u are so smart.
    please note that i have edited my post, i forgot about the second part of the question
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by helloying View Post
    the AC is not a straight line by the way. Just afraid my drawings are not clear
    Ooopss....
    I am very slow in typing. Let me post this anyway.



    Let AB be horizontal. So, BC is vertical.

    From D, drop a vertical line seggment to AB. Call the intersection as point E.
    From D also, draw a horizontal line segment to touch BC. Call the intersection as point F.

    ----------------------------
    1) To find lenght of BC:

    BC = BF +FC
    Since BF = DE, then, BC = DE +FC ----**

    In right triangle AED,
    sin(25deg) = DE / 9
    so, DE = 9sin(25deg) = 3.80356 units.

    In right triangle CFD,
    cos(40deg) = FC / 8
    So, FC = 8cos(40deg) = 6.12836 units

    therefore, BC = 3.80356 +6.12836 = 9.93192 units long. ------answer.

    ------------------------
    To find angle CAD:

    In the figure drawn above,
    >>>extend ED upwards some more, up to point G ...above hypotenuse AC.
    >>>extend FD to the left some more, up to point H ... beyond AC.

    angle FCV = angle GDC = 40 degrees
    angle EAD = angle HDA = 25 degrees
    angle GDH = 90 degrees
    so, angle CAD = 40 +90 +25 = 155 degrees. ----answer.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: May 9th 2011, 12:52 PM
  2. Tricky question to set up the DE for... probably easy to solve
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: August 2nd 2010, 10:35 AM
  3. Help me with my grade 11 maths [easy maths]
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: December 27th 2009, 04:09 PM
  4. I can't solve this easy question ?
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 8th 2009, 02:57 AM
  5. solve Polynominal question - easy
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: July 23rd 2009, 12:49 AM

Search Tags


/mathhelpforum @mathhelpforum