Results 1 to 4 of 4

Math Help - Working with Trig identies to show that a triangle is a right triangle

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    2

    Working with Trig identies to show that a triangle is a right triangle

    Can someone please assist me with this questions? Thank you in advance for your assistance.

    If angles A, B, and C are the angles of a triangle such that sin(A+B)=1/sin(C) and cos(A+B)=cos(C), then show that the triangle is a right triangle
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2009
    From
    Virginia
    Posts
    11
    Quote Originally Posted by bbanbury View Post
    Can someone please assist me with this questions? Thank you in advance for your assistance.

    If angles A, B, and C are the angles of a triangle such that sin(A+B)=1/sin(C) and cos(A+B)=cos(C), then show that the triangle is a right triangle
    Hmm, interesting! I'm sure there are many ways to prove that ABC is a right triangle, but here's one method:

    So let's assume ABC is a right triangle.

    c is the hypotenuse of the triangle
    a and b are both legs of the triangle
    Leg a is the opposite side to angle A; leg a isn't adjacent to angle A. So the sine of A would be \frac{a}{c}.

    cos(A+B) = cosC can be expanded to

    cos(A)cos(B) - sin(A)sin(B) = cosC

    \frac{b}{c} \times \frac{a}{c}-\frac{a}{c} \times \frac{b}{c} = cosC

    \frac{ab}{c^2} - \frac{ab}{c^2} = cosC

    \frac{ab}{c^2} - \frac{ab}{c^2} = 0

    cosC = 0

    The cosine of C is only 0 if angle C is 90^{\circ} or 270^{\circ}. Since a triangle cannot have a 270^{\circ} angle, angle C must be 90^{\circ}, and ABC is a right triangle.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570

    Trigonometry

    Hello bbanbury
    Quote Originally Posted by bbanbury View Post
    Can someone please assist me with this questions? Thank you in advance for your assistance.

    If angles A, B, and C are the angles of a triangle such that sin(A+B)=1/sin(C) and cos(A+B)=cos(C), then show that the triangle is a right triangle
    This is very straightforward.

    We know that, for any angle \theta,\, \sin\theta = \sin(180-\theta). And in a triangle ABC, A+B = 180 - C, so \sin(A+B) = \sin(180-C) = \sin C in any triangle.

    If, in addition, we know that \sin(A+B) = \frac{1}{\sin C}, then

    \sin C = \frac{1}{\sin C}

    \Rightarrow \sin^2C= 1

    \sin C = +1 (since \sin C = -1 is impossible if 0<C<180)

    \Rightarrow C = 90^o

    So the triangle is right-angled at C.

    (Notice that we don't need the extra fact about \cos(A+B) = \cos C. But you could equally well use this instead, and say that \cos C = -\cos(180 - C) = -\cos(A+B) in any triangle. So in this case \cos C = \cos(A+B) \Rightarrow \cos C = -\cos C = 0 \Rightarrow C=90^o.)

    Grandad
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2009
    Posts
    2

    A big Thank you

    Thank you for your assistance...it makes sense. Have a great day.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Triangle (Show)
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: January 14th 2010, 05:03 AM
  2. Working out height of a isosceles triangle
    Posted in the Geometry Forum
    Replies: 4
    Last Post: December 16th 2009, 01:01 PM
  3. Use the triangle inequality to show...
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: October 22nd 2009, 11:06 PM
  4. Replies: 1
    Last Post: October 28th 2008, 07:02 PM
  5. Replies: 27
    Last Post: April 27th 2008, 10:36 AM

Search Tags


/mathhelpforum @mathhelpforum