त्रिभुज कैलकुलेटर: SSS, SAS, ASA, AAS और SSA

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.

तीन मापों से त्रिभुज की भुजाएँ, कोण, क्षेत्रफल, परिमाप, ऊँचाइयाँ, अंतःत्रिज्या और परित्रिज्या निकालें। SSA के दोनों संभव हल देखें।

अपडेट किया गया 28 सितंबर 2026जाँचे गए उदाहरण: 8

क्षेत्रफल

cm²

अधिकतम दशमलव स्थान: 4; निकटतम; बराबर दूरी पर शून्य से दूर

Side a: 7cm
Side b: 8cm
Side c: 9cm
Angle A: 48.1897°
Angle B: 58.4119°
Angle C: 73.3985°
परिमाप: 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

गणना कैसे होती है S

Law of cosines for each angle
$\begin{gathered} \cos A = \frac{b^2 + c^2 - a^2}{2bc} = 0.666667 \Rightarrow A = 48.189685^\circ \\[6pt] \cos B = \frac{a^2 + c^2 - b^2}{2ac} = 0.52381 \Rightarrow B = 58.411864^\circ \\[6pt] \cos C = \frac{a^2 + b^2 - c^2}{2ab} = 0.285714 \Rightarrow C = 73.39845^\circ \end{gathered}$
क्षेत्रफल
$\text{Area} = \tfrac{1}{2}ab\sin C = \tfrac{1}{2} \times 7 \times 8 \times \sin 73.39845^\circ = 26.832816\,\text{cm}^{2}$
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
$\begin{gathered} h_a = \frac{2\,\text{Area}}{a} = 7.666519\,\text{cm} \\[6pt] h_b = \frac{2\,\text{Area}}{b} = 6.708204\,\text{cm},\quad h_c = \frac{2\,\text{Area}}{c} = 5.962848\,\text{cm} \end{gathered}$

त्रिभुज कैलकुलेटर: SSS, SAS, ASA, AAS और SSA के बारे में

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.

हल किए गए उदाहरण

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
परिणाम की इकाई: सेंटीमीटर (cm)

क्षेत्रफल: 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

जाँच का स्रोत: 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
परिणाम की इकाई: सेंटीमीटर (cm)

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

जाँच का स्रोत: 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 °
परिणाम की इकाई: सेंटीमीटर (cm)

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

जाँच का स्रोत: 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 °
परिणाम की इकाई: सेंटीमीटर (cm)

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

जाँच का स्रोत: Python 3.8 math: C = 80°, law of sines a = 10·sin40/sin80

सवाल

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.

“त्रिभुज कैलकुलेटर: SSS, SAS, ASA, AAS और SSA” कितना सटीक है?

सटीकता आपके इनपुट और विधि की मान्यताओं पर निर्भर करती है। दशमलव गणना 50 सार्थक अंकों का उपयोग करती है, लेकिन अनुमान, संख्यात्मक विधियाँ और स्रोत डेटा कम सटीक हो सकते हैं। दिखाए गए मानों को पूर्णांकित करने से ये सीमाएँ दूर नहीं होतीं। स्वतंत्र स्रोतों के हल किए गए उदाहरणों से जाँच: 8। उदाहरण के लिए, “SSS 3-4-5” की जाँच 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)) से की गई है।

इस विधि का स्रोत क्या है?

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

इस कैलकुलेटर के बारे में

\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}

स्रोत

संदर्भों से जाँच की गई

इस कैलकुलेटर में स्वतंत्र स्रोतों के उत्तरों वाले 8 हल किए गए उदाहरण शामिल हैं। ये परीक्षण समूह में चलते हैं और आप इन्हें यहाँ भी चला सकते हैं।