Consider triangle ABC and XYZ, with AB = XY, BC = YZ, and angle BAC = angle YXZ. Two sides and a non-included angle of one triangle are congruent to two sides and a non-included angle of the other triangle. Is it correct to conclude that Triangle ABC must be congruent to Trianalge XYZ?
Be carefull, I think this might be a trick question;
thanks,
-Dani
2007-03-09 15:46:37 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
hi , the answer is no for the given information , they are not congruent or the given data is not enought to prove congruence , cuz for two trinagles to be congruent there must be side , included angle ,side in one of them congruent to their corresponding elements in the other
2007-03-10 08:31:47 · answer #1 · answered by emy 3 · 0⤊ 0⤋
Well, in general there is no such triangle congruence for 2 pairs of congruent sides and a pair of non-included angles. In trigo, this is what we call the ambiguous case for sine law. However, there is an exception. If the angle is either a right or an obtuse angle, then there would be a congruence between the 2 triangles.
2007-03-09 23:55:31 · answer #2 · answered by Moja1981 5 · 0⤊ 0⤋
It's a trick question. Consider 2 triangles which are similiar but of different sizes. Both triangles will have all the same angles in the same places, and all the sides are even in the same proportion. But the LONGEST side of triangle 1 could be the same as the SHORTEST side of triangle 2. See? The answer is "no, you cannot conclude that"
2007-03-10 00:33:38 · answer #3 · answered by Scythian1950 7 · 0⤊ 0⤋