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

- Apr 29th 2009, 05:47 PMbbanburyWorking 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 - Apr 29th 2009, 06:32 PMlisczz
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.

$\displaystyle c$ is the hypotenuse of the triangle

$\displaystyle a$ and $\displaystyle b$ are both legs of the triangle

Leg $\displaystyle a$ is the opposite side to angle $\displaystyle A$; leg $\displaystyle a$ isn't adjacent to angle $\displaystyle A$. So the sine of A would be $\displaystyle \frac{a}{c}$.

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

$\displaystyle cos(A)cos(B) - sin(A)sin(B) = cosC$

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

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

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

$\displaystyle cosC = 0$

The cosine of $\displaystyle C$ is only $\displaystyle 0$ if angle $\displaystyle C$ is $\displaystyle 90^{\circ}$ or $\displaystyle 270^{\circ}.$ Since a triangle cannot have a $\displaystyle 270^{\circ}$ angle, angle $\displaystyle C$ must be $\displaystyle 90^{\circ}$, and ABC is a right triangle. - Apr 30th 2009, 05:33 AMGrandadTrigonometry
Hello bbanburyThis is very straightforward.

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

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

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

$\displaystyle \Rightarrow \sin^2C= 1$

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

$\displaystyle \Rightarrow C = 90^o$

So the triangle is right-angled at C.

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

Grandad - Apr 30th 2009, 07:41 AMbbanburyA big Thank you
Thank you for your assistance...it makes sense. Have a great day.