LoyalKnight
New member
- Joined
- Nov 14, 2013
- Messages
- 7
Problem:
Show that, in any triangle, we have the following:
. . . . .\(\displaystyle \dfrac{a\sin(A)\, +\, b\sin(B)\, +\, c\sin(C)}{a\cos(A)\, +\, b\cos(B)\, +\, c\cos(C)}\, =\, R\left(\dfrac{a^2\, +\, b^2\, +\, c^2}{abc}\right)\)
...where \(\displaystyle R\) is the circumradius of the triangle.
Problem hints:
i) Turn the problem into a trigonometry problem: get rid of the side lengths.
ii) Remember that \(\displaystyle A\, +\, B\, +\, C\, =\, 180^{\circ}\) for any triangle, so \(\displaystyle C\, =\, 180^{\circ}\, -\, \left(A\, +\, B\right).\)
How to solve this? I can't figure out how. Help is appreciated, thanks.
Show that, in any triangle, we have the following:
. . . . .\(\displaystyle \dfrac{a\sin(A)\, +\, b\sin(B)\, +\, c\sin(C)}{a\cos(A)\, +\, b\cos(B)\, +\, c\cos(C)}\, =\, R\left(\dfrac{a^2\, +\, b^2\, +\, c^2}{abc}\right)\)
...where \(\displaystyle R\) is the circumradius of the triangle.
Problem hints:
i) Turn the problem into a trigonometry problem: get rid of the side lengths.
ii) Remember that \(\displaystyle A\, +\, B\, +\, C\, =\, 180^{\circ}\) for any triangle, so \(\displaystyle C\, =\, 180^{\circ}\, -\, \left(A\, +\, B\right).\)
How to solve this? I can't figure out how. Help is appreciated, thanks.
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