Results 1 to 9 of 9

Math Help - Triangle (Proving)

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    78

    Triangle (Proving)

    I'm getting very confused...there are too many formula for this!

    Prove:
    1. sin (A+B) + cos (A-B) \equiv (sin A+ cos A)(sin B + cos B)

    2. tan (A+B) - tan A \equiv \frac{sin B}{cos A cos (A+B)}

    And one more extra:
    3. In triangle ABC, AB=p, BC=3p, and \angle ABC=60^o. The bisector of the angle ABC meets AC at the point T. Find AT and CT in terms of p.

    Answer: AT=\frac{p}{4}\sqrt{7}; CT=\frac{3p}{4}\sqrt{7}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by cloud5 View Post
    I'm getting very confused...there are too many formula for this!

    Prove:
    1. sin (A+B) + cos (A-B) \equiv (sin A+ cos A)(sin B + cos B)

    2. tan (A+B) - tan A \equiv \frac{sin B}{cos A cos (A+B)}

    And one more extra:
    3. In triangle ABC, AB=p, BC=3p, and \angle ABC=60^o. The bisector of the angle ABC meets AC at the point T. Find AT and CT in terms of p.

    Answer: AT=\frac{p}{4}\sqrt{7}; CT=\frac{3p}{4}\sqrt{7}
    HI

    (1) \sin A\cos B+\cos A\sin B+\cos A\cos B+\sin A\sin B

    \sin A(\cos B+\sin B)+\cos A(\cos B+\sin B)

    (\sin A+\cos A)(\sin B+\cos B)

    (2) \frac{\sin (A+B)}{\cos (A+B)}-\frac{\sin A}{\cos A}

    \frac{\sin (A+B)\cos A-\sin A\cos (A+B)}{\cos A\cos (A+B)}

    \frac{\sin A\cos B\cos A+\cos A\sin B\cos A-\cos A\cos B\sin A+\sin^2 A\sin B}{\cos A\cos (A+B)}

    \frac{(\cos^2 A+\sin^2 A)\sin B}{\cos A\cos (A+B)}

    so the result follow .

    (3) Make a sketch first . The ratio of AT:TC=1:3

    By the cosine rule , AC^2=p^2+(3p)^2-2p(3p)\cos 60

    AC=p\sqrt{7}

    AT=\frac{1}{4}AC=\frac{1}{4}p\sqrt{7}

    sure you can find TC now .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2009
    Posts
    78
    Quote Originally Posted by mathaddict View Post
    \frac{\sin A\cos B\cos A+\cos A\sin B\cos A-\cos A\cos B\sin A+\sin^2 A\sin B}{\cos A\cos (A+B)}

    \frac{(\cos^2 A+\sin^2 A)\sin B}{\cos A\cos (A+B)}

    so the result follow .
    Can you explain what happen between those steps?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by cloud5 View Post
    Can you explain what happen between those steps?
    sure , notice that \cos A\cos B\sin A cancels off and you are left with

    \frac{\cos^2 A\sin B+\sin^2 A\sin B}{\cos A\cos (A+B)}

    then bring out the common factor , sin B .
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jun 2009
    Posts
    78
    Quote Originally Posted by mathaddict View Post

    AC=p\sqrt{7}

    AT=\frac{1}{4}AC=\frac{1}{4}p\sqrt{7}

    sure you can find TC now .
    I don't understand this part... how you got \sqrt{7}?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by cloud5 View Post
    I don't understand this part... how you got \sqrt{7}?
    Note that cos 60=1/2

    AC^2=10p^2-6p^2(\frac{1}{2})

     <br />
AC^2=7p^2 so AC=p\sqrt{7}<br />
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jun 2009
    Posts
    78
    Quote Originally Posted by mathaddict View Post

    AT=\frac{1}{4}AC=\frac{1}{4}p\sqrt{7}
    One more question, how you get \frac{1}{4}?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by cloud5 View Post
    One more question, how you get \frac{1}{4}?
    because AT:TC is in the ratio of 1:3 , the 'total' for AC is 4 .

    so AT = 1/4 AC
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jun 2009
    Posts
    78
    Quote Originally Posted by mathaddict View Post
    because AT:TC is in the ratio of 1:3 , the 'total' for AC is 4 .

    so AT = 1/4 AC
    How to see the ratio(it dosen't state in the question)? Is it from the sketch? Sorry I know I'm asking too much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving a triangle in a circle is degenerate
    Posted in the Geometry Forum
    Replies: 2
    Last Post: July 6th 2011, 04:05 AM
  2. Proving a triangle is right-angled.
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: April 26th 2010, 01:26 PM
  3. Proving Triangle Sides
    Posted in the Geometry Forum
    Replies: 2
    Last Post: February 2nd 2010, 03:20 PM
  4. Proving Right-Angled/Right Angle Triangle Help
    Posted in the Geometry Forum
    Replies: 3
    Last Post: January 13th 2010, 05:57 AM
  5. proving a right triangle
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 15th 2007, 08:21 PM

Search Tags


/mathhelpforum @mathhelpforum