حاسبة المثلثات: SSS وSAS وASA وAAS وSSA

Q: How do you find the angles of a triangle from its three sides?

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.

Q: How do you find the area of a triangle without the height?

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.

Q: How can you tell if a triangle is acute, right or obtuse from its sides?

Compare the square of the longest side c with the sum of the squares of the other two. If c² 41, obtuse.

أوجد أضلاع المثلث وزواياه ومساحته ومحيطه وارتفاعاته ونصفي قطر الدائرتين الداخلية والمحيطة من ثلاثة قياسات، مع حلي SSA الممكنين.

آخر تحديث 28 سبتمبر 2026أمثلة تم التحقق منها: 8

المساحة

cm²

الحد الأقصى للمنازل العشرية: 4؛ إلى الأقرب، وعند التعادل بعيدًا عن الصفر

Side a: 7cm
Side b: 8cm
Side c: 9cm
Angle A: 48.1897°
Angle B: 58.4119°
Angle C: 73.3985°
المحيط: 24cm
Inradius: 2.2361cm
Circumradius: 4.6957cm
Height onto a: 7.6665cm
Height onto b: 6.7082cm
Height onto c: 5.9628cm
Type: Scalene, acute
Number of triangles: 1

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.

The triangle, to scale

طريقة الحساب S

Law of cosines for each angle
$\begin{gathered} \cos A = \frac{b^2 + c^2 - a^2}{2bc} = 0.666667 \Rightarrow A = 48.189685^\circ \\[6pt] \cos B = \frac{a^2 + c^2 - b^2}{2ac} = 0.52381 \Rightarrow B = 58.411864^\circ \\[6pt] \cos C = \frac{a^2 + b^2 - c^2}{2ab} = 0.285714 \Rightarrow C = 73.39845^\circ \end{gathered}$
المساحة
$\text{Area} = \tfrac{1}{2}ab\sin C = \tfrac{1}{2} \times 7 \times 8 \times \sin 73.39845^\circ = 26.832816\,\text{cm}^{2}$
Inradius and circumradius
$\begin{gathered} r = \frac{\text{Area}}{s} = \frac{26.832816}{12} = 2.236068\,\text{cm} \\[6pt] R = \frac{a}{2\sin A} = 4.695743\,\text{cm} \end{gathered}$
s is the semi-perimeter.
Heights
$\begin{gathered} h_a = \frac{2\,\text{Area}}{a} = 7.666519\,\text{cm} \\[6pt] h_b = \frac{2\,\text{Area}}{b} = 6.708204\,\text{cm},\quad h_c = \frac{2\,\text{Area}}{c} = 5.962848\,\text{cm} \end{gathered}$

حول حاسبة المثلثات: SSS وSAS وASA وAAS وSSA

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.

أمثلة محلولة

SSS 3-4-5

I know: SSS
Side a (opposite A): 3 cm
Side b (opposite B): 4 cm
Side c (opposite C): 5 cm
وحدة النتائج: سنتيمتر (⁨cm⁩)

المساحة: 6 cm²
Angle C: 90 °
Angle A: 36.869898 °
Angle B: 53.130102 °
Inradius: 1 cm
Circumradius: 2.5 cm
Type: Scalene, right
Number of triangles: 1

مصدر التحقق: ⁨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))⁩

Equilateral, side 5

I know: SSS
Side a (opposite A): 5 cm
Side b (opposite B): 5 cm
Side c (opposite C): 5 cm
وحدة النتائج: سنتيمتر (⁨cm⁩)

المساحة: 10.825318 cm²
Angle A: 60 °
Angle B: 60 °
Angle C: 60 °
Type: Equilateral
Circumradius: 2.886751 cm

مصدر التحقق: ⁨Python 3.8 math: sqrt(3)/4*25, 5/sqrt(3)⁩

SAS a = 5, b = 7, C = 49°

I know: SAS
Side a (opposite A): 5 cm
Side b (opposite B): 7 cm
Angle C (between a and b): 49 °
وحدة النتائج: سنتيمتر (⁨cm⁩)

Side c: 5.298667 cm
Angle A: 45.411694 °
Angle B: 85.588306 °
المساحة: 13.207418 cm²

مصدر التحقق: ⁨Python 3.8 math: law of cosines and ½ab·sin C⁩

ASA A = 40°, c = 10, B = 60°

I know: ASA
Side c (opposite C): 10 cm
Angle A: 40 °
Angle B: 60 °
وحدة النتائج: سنتيمتر (⁨cm⁩)

Angle C: 80 °
Side a: 6.527036 cm
Side b: 8.793852 cm
المساحة: 28.262897 cm²

مصدر التحقق: ⁨Python 3.8 math: C = 80°, law of sines a = 10·sin40/sin80⁩

الأسئلة

How do you find a missing side of a triangle that is not right-angled?

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.

What is the ambiguous case of the law of sines?

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°.

How do you find the angles of a triangle from its three sides?

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.

How do you find the area of a triangle without the height?

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.

How can you tell if a triangle is acute, right or obtuse from its sides?

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.

ما مدى دقة «⁨حاسبة المثلثات: SSS وSAS وASA وAAS وSSA⁩»؟

تعتمد الدقة على مدخلاتك وافتراضات الطريقة. يستخدم الحساب العشري 50 رقمًا معنويًا، لكن التقديرات والأساليب العددية وبيانات المصدر قد تكون أقل دقة؛ تقريب القيم المعروضة لا يزيل هذه الحدود. أمثلة محلولة جرى التحقق منها بمصادر مستقلة: 8. مثلًا، يجري التحقق من «⁨SSS 3-4-5⁩» بالرجوع إلى ⁨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))⁩.

ما مصدر هذه الطريقة؟

Weisstein, E. W. “Law of Sines”, “Law of Cosines”, “Triangle” — MathWorld; OpenStax Precalculus 2e, §8.1 Non-right Triangles: Law of Sines (ambiguous case).

حول هذه الحاسبة

\begin{gathered} \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,\qquad c^2 = a^2 + b^2 - 2ab\cos C \\ \text{Area} = \tfrac{1}{2}ab\sin C,\qquad r = \frac{\text{Area}}{s} \end{gathered}

المصادر

تم التحقق بالرجوع إلى المصادر

تتضمن هذه الحاسبة أمثلة محلولة بإجابات من مصادر مستقلة، وعددها 8. تُشغّل ضمن مجموعة الاختبارات، ويمكنك تشغيلها هنا أيضًا.