Q: ABC and DEF are angles.

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If the sides AB, BC and CA of triangle ABC correspond to the sides DE, EF and FD of triangle DEF, then the two triangles are congruent if:AB = DE, BC = EF and CA = FD (SSS)AB = DE, BC = EF and angle ABC = angle DEF (SAS)AB = DE, angle ABC = angle DEF, angle BCA = angle EFD (ASA)If the triangles are right angled at A and D so that BC and EF are hypotenuses, then the triangles are congruent ifBC = EF and AB = DE (RHS)BC = EF and angle ABC = angle DEF (RHA).

The sum of the two angles is 360. So angle ABC = 120 degrees.

Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc. There is, therefore, no visible symbol between ABC and DEF (<, =, >, ≠ etc). Furthermore, there is no information as to whether ABC is an angle, a triangle, an arc.

Nothing else, the angle-angle-side is sufficient to show the triangles are congruent. With two corresponding angles are equal, the third angles in the triangles by definition (the sum of the three angles in a triangle is 180o) must be equal making the triangles similar. If a corresponding pair of sides are also equal, then the other two corresponding pairs of sides will be equal.

Angle abc will form a right angle if and only if, segment ab is perpendicular to segment bc.

Related questions

They are congruent when they have 3 identical dimensions and 3 identical interior angles.

It depends on where and what ABC and DEF are!

4,8,12

false

Answer: Since you are looking for the scale factor of ABC to DEF the answer is 8 because DEF is 8 times larger than ABC.

false

Transitive

ABC

False. If ABC definitely equals DEF equals MNO and MNO equals PQR then ABC does not equal PQR by the transitive property.

You can declare and implement a class within a class...class abc {...class def {...}}The class def is scoped and known only within the context of abc. It is possible to declare instances of def as member variables of abc, however, for this to work correctly, they should be private, because methods of def, not even public methods, do not exist outside of the scope of abc.

Yes because adg beh cfi is just abc def ghi mixed up.

They are 17 times AB, BC and Ca, respectively.