세 측정값으로 변, 각, 넓이, 둘레, 높이, 내접원과 외접원의 반지름을 구하고 SSA의 가능한 두 해를 확인하세요.
업데이트 검증한 예제: 8
The triangle has sides 7, 8 and 9 cm with angles 48.1897°, 58.4119° and 73.3985°; it is scalene, acute and encloses 26.8328 cm².
s is the semi-perimeter.
Three measurements fix a triangle as long as at least one of them is a side. The law of cosines, c² = a² + b² − 2ab cos C, solves three sides (SSS) and two sides with the angle between them (SAS). The law of sines, a/sin A = b/sin B = c/sin C, solves two angles and a side (ASA, AAS). The area comes from ½ab sin C, and the results add the perimeter, all three heights, the inradius and the circumradius.
The default sides of 7, 8 and 9 cm give angles of 48.19°, 58.41° and 73.40° and an area of 26.8328 cm². Surveying and roof framing use the same two laws to find a length that cannot be measured directly from ones that can.
Two sides and an angle that is not between them (SSA) can fit two triangles, one or none. With a = 6, b = 8 and A = 35°, both B = 49.89° and B = 130.11° work, so both triangles are drawn and compared in a table.
검증 출처: Right triangle: area ½·3·4, r = (3 + 4 − 5)/2, R = hypotenuse/2; Python 3.8 math: degrees(acos(0.8)), degrees(acos(0.6))
검증 출처: Python 3.8 math: sqrt(3)/4*25, 5/sqrt(3)
검증 출처: Python 3.8 math: law of cosines and ½ab·sin C
검증 출처: Python 3.8 math: C = 80°, law of sines a = 10·sin40/sin80
With two sides and the angle between them, use the law of cosines: c² = a² + b² − 2ab cos C. For a = 5, b = 7 and C = 49°, c = √(74 − 70 cos 49°) ≈ 5.2987. With two angles and any side, use the law of sines, a/sin A = b/sin B. When C = 90° the cosine term is zero and the law of cosines becomes Pythagoras' theorem.
It is the SSA case: two sides and an angle opposite one of them can fit two different triangles. For an acute angle A, find the height h = b sin A. If a < h there is no triangle, if a = h one right triangle, if h < a < b two triangles, and if a ≥ b one. For a = 6, b = 8 and A = 35°, h = 4.589, so B is either 49.89° or 130.11°.
Rearrange the law of cosines: cos A = (b² + c² − a²)/(2bc), and the same pattern for B and C. For sides 7, 8 and 9, cos A = (64 + 81 − 49)/144 = 0.6667, so A = 48.19°; B = 58.41° and C = 73.40° follow the same way, and the three add up to 180°. The largest angle is always opposite the longest side.
Take half the product of two sides and the sine of the angle between them: Area = ½ab sin C. Sides 7 and 8 with C = 73.40° give ½ × 7 × 8 × 0.9583 ≈ 26.83. From three sides alone, Heron's formula gives the same answer: the semi-perimeter is 12, and √(12 × 5 × 4 × 3) = √720 ≈ 26.83.
Compare the square of the longest side c with the sum of the squares of the other two. If c² < a² + b² the triangle is acute, if they are equal it has a right angle, and if c² is larger it is obtuse. Sides 7, 8 and 9 give 81 < 113, so acute; 3, 4 and 5 give 25 = 25, right; 4, 5 and 8 give 64 > 41, obtuse.
정확도는 입력값과 계산 방법의 가정에 따라 달라집니다. 십진 연산은 유효숫자 50자리를 사용하지만, 추정값·수치해석 방법·원본 데이터의 정밀도는 더 낮을 수 있습니다. 표시값을 반올림해도 이러한 한계는 사라지지 않습니다. 독립적인 출처의 풀이와 대조한 계산 예시: 8. 예를 들어 “SSS 3-4-5”은 Right triangle: area ½·3·4, r = (3 + 4 − 5)/2, R = hypotenuse/2; Python 3.8 math: degrees(acos(0.8)), degrees(acos(0.6))와 대조해 확인합니다.
Weisstein, E. W. “Law of Sines”, “Law of Cosines”, “Triangle” — MathWorld; OpenStax Precalculus 2e, §8.1 Non-right Triangles: Law of Sines (ambiguous case).
이 계산기에는 독립적인 출처에서 답을 얻은 계산 예제가 8개 있습니다. 테스트 모음에서 실행되며 여기에서도 실행할 수 있습니다.