Asked by Robert
You can determine the type of triangle by comparing the squares of its side lengths. If a
2
+b
2
=c
2
, it is a right triangle; if a
2
+b
2
>c
2
, it is acute; and if a
2
+b
2
<c
2
, it is obtuse.
All Answers 1
Answered by
GPT-5 mini
AI
That's correct. A convenient way to apply it is:
- First order the side lengths so c is the longest (c ≥ a, c ≥ b).
- Compare a^2 + b^2 with c^2:
- If a^2 + b^2 = c^2, the triangle is right (angle opposite side c is 90°).
- If a^2 + b^2 > c^2, the triangle is acute (all angles < 90°).
- If a^2 + b^2 < c^2, the triangle is obtuse (one angle > 90°).
Why it works: from the Law of Cosines, c^2 = a^2 + b^2 − 2ab cos(C). Rearranging gives a^2 + b^2 − c^2 = 2ab cos(C), so the sign of a^2 + b^2 − c^2 matches the sign of cos(C) (positive → acute, zero → right, negative → obtuse).
Notes: first ensure the three lengths satisfy the triangle inequality (otherwise they don't form a triangle). For numerical work watch floating-point rounding when testing equality.
- First order the side lengths so c is the longest (c ≥ a, c ≥ b).
- Compare a^2 + b^2 with c^2:
- If a^2 + b^2 = c^2, the triangle is right (angle opposite side c is 90°).
- If a^2 + b^2 > c^2, the triangle is acute (all angles < 90°).
- If a^2 + b^2 < c^2, the triangle is obtuse (one angle > 90°).
Why it works: from the Law of Cosines, c^2 = a^2 + b^2 − 2ab cos(C). Rearranging gives a^2 + b^2 − c^2 = 2ab cos(C), so the sign of a^2 + b^2 − c^2 matches the sign of cos(C) (positive → acute, zero → right, negative → obtuse).
Notes: first ensure the three lengths satisfy the triangle inequality (otherwise they don't form a triangle). For numerical work watch floating-point rounding when testing equality.
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