How to Use the Triangle Calculator
- Select the calculation mode.
- Enter the required calculation data.
- Click the “Calculate” button.
- View the example diagram and calculation results on the right.
- Click the “Share” button at the bottom left to share your calculation results.
Given the three side lengths of triangle \(△ABC\), calculate its other mathematical properties
Given the three side lengths \(a\), \(b\), and \(c\) of triangle \(△ABC\) (where \(a\) corresponds to side \(BC\), \(b\) to side \(AC\), and \(c\) to side \(AB\)), the following properties can be calculated:
perimeter, area, interior angles, and circumference. \(c\) (where \(a\) corresponds to side \(BC\), \(b\) to side \(AC\), and \(c\) to side \(AB\)), we can calculate properties such as perimeter, area, interior angles, radius of the circumcircle, and radius of the incircle. The specific methods are as follows:
1. How to calculate the perimeter \(L\)of a triangle?
The perimeter is the sum of the three side lengths, given by the formula:
2. How to calculate the area \(S\)of a triangle?
To calculate the area of a triangle, we can use Heron’s formula. The steps are as follows:
- Step 1: Calculate the semi-perimeter \( p\) of the triangle.
- Step 2: Substitute the result from Step 1 into Heron’s formula to find the area.
3. How to calculate the three interior angles \( ∠A \), \( ∠B \), \( ∠C \)of a triangle?
To calculate the three interior angles of a triangle, we can use the cosine rule to find the cosine value of each angle, then use the inverse cosine function to determine the angle (result in radians or degrees, convert as needed):
Verification: The sum of the interior angles of a triangle is \( 180^\circ \) (or \(\ pi \) radians), i.e., \( ∠A+∠B+∠C=180^\circ \).
4. How to calculate the radius \(R\)of a triangle’s circumcircle?
The circumcircle of a triangle is a circle passing through all three vertices. Its radius formula is derived from the sine rule:
A more direct formula can also be derived using the area formula:
5. How to calculate the radius \( r \)of a triangle’s incircle?
The incircle is a circle tangent to all three sides of the triangle. Its radius formula is:
where \( S \) is the triangle’s area and \( p \) is its semi-perimeter.
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Example Calculation
Assume triangle \(△ABC\) has side lengths \(a=30\), \(b=40\), \(c=50\)
1. Semicircumference \(p=\frac{30+40+50}{2}=60\)
2. Area \(S = \sqrt{60 \times (60 – 30) \times (60 – 40) \times (60 – 50)} = \sqrt{60 \times 30 \times 20 \times 10} = \sqrt{360000} = 600\)
3. Interior angle \(∠C\): \(\cos C = \frac{30^2 + 40^2 – 50^2}{2 \times 30 \times 40} = \frac{900 + 1600 – 2500}{2400} = 0\), so \(∠C = 90^\circ\)
4. Radius of the circumcircle \(R=\frac{30\times40\times50}{4\times600}=\frac{60000}{2400}=25\)
5. Radius of the incircle \(r=\frac{600}{60}=10\)
Given the lengths of two sides (sides a and b corresponding to angles A and B) and the measure of the included angle C (∠ACB) in triangle △ABC, the other mathematical properties of this triangle
Given triangle \(△ABC\), where side \(a\) (corresponding to angle \(∠A\), i.e., side \(BC\)) and side \(b\) (corresponding to angle \(∠B\), i.e., side \(AC\)) are known, along with the angle \(∠C\) (i.e., the angle between \(AC\) and \(BC\)), the following properties can be calculated sequentially: the third side \(c\), the perimeter \(L\), the area \(S\), the other two interior angles \(∠A\) and \(∠B\), the radius of the circumcircle \(R\), and the radius of the incircle \(r\). The specific calculation methods are as follows:
1. How to calculate the third side \(c\)?
Finding the third side given two sides and the included angle is a core application of the Law of Cosines. Substitute directly into the formula:
Since side lengths are positive, take the square root (positive root) of the above result to obtain the third side length:
Note: Ensure the unit of \(∠C\) matches the trigonometric calculation requirements (if using degrees, set the calculator to degree mode; if using radians, set it to radian mode).
2. How to calculate the perimeter \(L\)of a triangle?
The perimeter is the sum of the three side lengths. Since the third side \(c\) was found in Step 1, the perimeter formula remains the same as before:
You can either substitute \(c\)’s expression to combine terms and calculate, or first find the exact value of \(c\) before summing.
3. How to calculate the area \(S\)of a triangle?
When two sides and the included angle are known, prioritize using the area formula for a triangle with two sides and an angle. This eliminates the need to calculate the third side, simplifying the steps. The formula is:
Additional note: Alternatively, you can first find the third side \(c\) and then use Heron’s formula to calculate the area (first calculate the semi-perimeter \(p=\frac{a+b+c}{2}\), then substitute into \(S=\sqrt{p(p-a)(p-b)(p-c)}\)). Both methods yield the same result, but the angle-between-two-sides formula is more efficient.
4. How to calculate the other two interior angles \(∠A\)and \(∠B\)?
They can be determined using the Law of Sines. This theorem establishes proportional relationships between sides and opposite angles. Combined with the fact that the sum of a triangle’s interior angles is \(180^\circ\) (\(\pi\) radians), the two unknown angles can be quickly calculated. The specific steps and formula are as follows:
Core formula of the Law of Sines:
Step 1: Find \( ∠A \) (the angle corresponding to side \( a \)). Rearrange the Law of Sines to obtain:
Calculate the angle using the inverse sine function: \(∠A = \arcsin\left(\frac{a \cdot \sin C}{c}\right)\) (Note: If \(a \geq c\), verify whether \(∠A\) is obtuse to avoid multiple solutions).
Step 2: Find ∠B (the angle corresponding to side \(b\)) using the triangle angle sum theorem:
- Step 1: Find one angle (e.g., ∠A) using the sine rule
- Step 2: Find another angle (e.g., \(∠B\)) using the triangle’s interior angle sum
5. How to calculate the radius \(R\)of a triangle’s circumcircle?
This can still be derived using the Law of Sines. Another form of the Law of Sines directly relates the side lengths, opposite angles, and the radius of the circumcircle. The formula is:
Simply substitute any pair of “side length-corresponding angle” (e.g., known \(a\) and calculated \(∠A\), or \(b\) and \(∠B\), or \(c\) and \(∠C\)) to compute the result, which will be consistent. Alternatively, use the area formula with the derived expression: \(R = \frac{abc}{4S}\) (requires finding \(c\) and \(S\) first).
6. How to calculate the radius \( r \)of a triangle’s incircle?
The incircle radius formula aligns with previous discussions, fundamentally relating area \( S \) and semi-perimeter \( p \):
Additional note: The semi-perimeter \(p = \frac{a + b + c}{2}\), and the area \(S\) can be calculated using the angle-between-two-sides formula (\(S = \frac{1}{2}ab\sin C\)), eliminating the need for Heron’s formula and simplifying the steps.
Where: \(S\) is the triangle’s area (calculated using either the angle-between-two-sides formula or Heron’s formula), \(p\) is the semi-perimeter, i.e., \(p = \frac{a + b + c}{2}\) (requiring the third side \(c\) to be determined first).
Example Calculation
Assume in triangle \(△ABC\), \(a=30\) (side \(BC\)), \(b=40\) (side \(AC\)), and the included angle \(∠C=90^\circ\). The specific calculation is as follows:
1. Find the third side \(c\):
Substitute into the cosine rule: \(c^2 = 30^2 + 40^2 – 2 \times 30 \times 40 \times \cos90^\circ\)
Given \(\cos90^\circ=0\), we calculate: \(c^2 = 900 + 1600 – 2400 \times 0 = 2500\)
Therefore \(c = \sqrt{2500} = 50\)
2. Find the perimeter \(L\):
\(L = a + b + c = 30 + 40 + 50 = 120\)
3. Find the area \(S\):
Using the angle-between-two-sides formula: \(S = \frac{1}{2} \times 30 \times 40 \times \sin90^\circ\)
Given \(\sin90^\circ = 1\), we calculate: \(S = 600 \times 1 = 600\)
Additional verification: Since ∠C = 90°, we can also directly use the area formula for a right triangle: \(S = \frac{1}{2}ab = 600\), yielding the same result.
4. Find the interior angles ∠A and ∠B:
① Find \(∠A\): By the Law of Sines, \(\sin A = \frac{a \cdot \sin C}{c} = \frac{30 \times \sin90^\circ}{50} = \frac{30 \times 1}{50} = 0.6\)
Therefore \(∠A = \arcsin(0.6) \approx 36.87^\circ\) (Verification: \(a=30 < c=50\), thus \(∠A < ∠C=90^\circ\), making it an acute angle with no multiple solutions)
② Find \(∠B\): \(∠B = 180^\circ – ∠A – ∠C \approx 180^\circ – 36.87^\circ – 90^\circ = 53.13^\circ\)
5. Find the radius \(R\)of the circumcircle:
Substitute \(a\) and \(\angle A\): \(R = \frac{a}{2\sin A} = \frac{30}{2 \times 0.6} = \frac{30}{1.2} = 25\)
Verification: Using the derived formula \(R = \frac{abc}{4S} = \frac{30 \times 40 \times 50}{4 \times 600} = \frac{60000}{2400} = 25\), consistent with the previous result.
6. Find the radius \(r\)of the inscribed circle:
First calculate the semi-perimeter \(p = \frac{a + b + c}{2} = \frac{120}{2} = 60\)
Then substitute into the formula: \(r = \frac{S}{p} = \frac{600}{60} = 10\)
Verification: The sum of interior angles \(36.87^\circ + 53.13^\circ + 90^\circ = 180^\circ\) satisfies the triangle angle sum theorem; The radius of the inscribed circle can also be verified using the special formula for right triangles: \(r = \frac{a + b – c}{2} = \frac{30 + 40 – 50}{2} = 10\), yielding consistent results.
Given the side A, angle B, and angle C of a triangle, calculate its other mathematical properties.
Given triangle \(△ABC\), where side \(a\) (corresponding to angle \(∠A\), i.e., side \(BC\)), angle \(∠B\) (corresponding to side \(b\) of \(AC\)), and angle \(∠C\) (corresponding to side \(c\) of \(AB\)) First, determine ∠A using the triangle’s interior angle sum. Then, apply the sine rule to find the other two sides \(b\) and \(c\). Subsequently, calculate the perimeter \(L\), area \(S\), circumcircle radius \(R\), and incircle radius \(r\). The specific calculation methods are as follows:
1. How to calculate the unknown interior angle \(∠A\)?
Based on the triangle interior angle sum theorem (the sum of the three interior angles of a triangle is \(180^\circ\), or \(\pi\) radians), directly derive and calculate using the formula:
Hint: If angles are given in radians, adjust the formula to \(∠A = \pi – ∠B – ∠C\) to ensure consistent units.
2. How to calculate unknown sides \(b\)and \(c\)?
Given one side \(a\) and its corresponding angle, along with the other two angles, prioritize solving for the unknown sides using the Law of Sines. The core principle of the Law of Sines is establishing fixed proportional relationships between sides and their opposite angles. The formula is:
By rearranging the formula, sides \(b\) and \(c\) can be calculated separately:
To find side \(b\) (corresponding to ∠B):
To find side \(c\) (corresponding to ∠C):
Note: Ensure trigonometric calculations use consistent angle units (degrees/radians) to avoid errors.
3. How to calculate the perimeter \(L\)of the triangle?
Perimeter is the sum of the three side lengths. Having determined sides \(b\) and \(c\) via the above steps, the formula remains consistent with previous sections:
You may substitute the expressions for \(b\) and \(c\) to simplify the calculation, or first determine the specific values of \(b\) and \(c\) before summing.
4. How to calculate the area \(S\)of a triangle?
Given one side and two angles, you can first use the Law of Sines to find any other side. Then, combine this with the area formula for a triangle with two sides and an included angle. This method is straightforward:
Method: First, use the Law of Sines to find side \(b\). Then, with sides \(a\) and \(b\) as known sides and the angle between them as \(∠C\), substitute into the area formula:
Additional note: Other combinations (e.g., sides \(a\) and \(c\) with angle \(∠B\), or sides \(b\) and \(c\) with angle \(∠A\)) may be used as long as the “two sides” are known/calculated lengths and the “angle” is the corresponding angle between them. The result remains consistent.
5. How to calculate the radius \(R\)of a triangle’s circumcircle?
Directly apply the derived formula from the Law of Sines. Given side \(a\) and its corresponding angle \(∠A\) (or any other “side-corresponding angle” combination), it can be calculated directly using the formula:
Verification using other “side-corresponding angle” pairs yields identical results: \(R = \frac{b}{2\sin B}\) and \(R = \frac{c}{2\sin C}\).
6. How to calculate the radius \(r\)of a triangle’s incircle?
The formula aligns with previous sections, fundamentally relating area \(S\) and semi-perimeter \(p\):
where the semi-perimeter \(p = \frac{a + b + c}{2} \) (requiring prior calculation of \( b \) and \( c \)), and the area \( S \) can be determined using the “area formula for a triangle between two sides” discussed earlier.
Example Calculation
Assume in triangle \(△ABC\), side \(a=30\) (side \(BC\), corresponding to \(∠A\)), \(∠B=60^\circ\), \(∠C=30^\circ\). The specific calculations are as follows:
1. Find the unknown interior angle \(∠A\):
Substitute into the interior angle sum formula: \(∠A = 180^\circ – 60^\circ – 30^\circ = 90^\circ\)
2. Find the unknown sides \(b\)and \(c\):
① Find side \(b\): Using the Law of Sines, \(b = \frac{a \cdot \sin B}{\sin A} = \frac{30 \cdot \sin60^\circ}{\sin90^\circ}\)
Given \(\sin60^\circ = \frac{\sqrt{3}}{2} \approx 0.866\) and \(\sin90^\circ = 1\), we calculate: \(b = 30 \times 0.866 \approx 25.98\)
② Find side \(c\): By the Law of Sines, \(c = \frac{a \cdot \sin C}{\sin A} = \frac{30 \cdot \sin30^\circ}{\sin90^\circ}\)
Given \(\sin30^\circ = 0.5\), we calculate: \(c = 30 \times 0.5 = 15\)
3. Find the perimeter \(L\):
\(L = a + b + c = 30 + 25.98 + 15 = 70.98\) (If retaining the square root, \(b = 15\sqrt{3}\), then \(L = 45 + 15\sqrt{3} \approx 70.98\))
4. Calculate the area \(S\):
Select sides \(a\), \(c\), and angle \(∠B\), then substitute into the formula: \(S = \frac{1}{2}ac\sin B\)
Calculate: \(S = \frac{1}{2} \times 30 \times 15 \times 0.866 \approx 194.85\) (with square root retained: \(S = \frac{225\sqrt{3}}{2} \approx 194.85\))
5. Find the radius \(R\)of the circumcircle:
Substitute side \(a\) and angle \(∠A\): \(R = \frac{a}{2\sin A} = \frac{30}{2 \times 1} = 15\)
Verification: Using side \(c\) and ∠C, \(R = \frac{c}{2\sin C} = \frac{15}{2 \times 0.5} = 15\), consistent results.
6. Find the radius \(r\)of the inscribed circle:
First calculate the semicircumference \(p = \frac{a + b + c}{2} = \frac{30 + 25.98 + 15}{2} \approx 35.49\)
Substitute into the formula: \(r = \frac{S}{p} = \frac{194.85}{35.49} \approx 5.49\) (calculate without the square root: \(p = \frac{45 + 15\sqrt{3}}{2}\)), \(r = \frac{\frac{225\sqrt{3}}{2}}{\frac{45 + 15\sqrt{3}}{2}} = \frac{225\sqrt{3}}{45 + 15\sqrt{3}} = \frac{15\sqrt{3} – 15}{2} \approx 5.49\)）
Verification: The sum of the interior angles \(90^\circ + 60^\circ + 30^\circ = 180^\circ\) satisfies the angle sum theorem; The radius of the circumcircle can also be verified using the special property of right triangles (the radius of a right triangle’s circumcircle is half the hypotenuse; here, hypotenuse \(a=30\), so \(R=15\)), yielding consistent results.
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Given sides \(a\)and \(b\)of a right triangle, calculate other geometric properties
Given right triangle \(△ABC\) with \(∠C=90^\circ\) (right angle), side \(a\) (corresponding to \(BC\), the right angle side), and side \(b\) (corresponding to \(AC\), the right angle side), Utilize the special properties of right triangles (Pythagorean theorem, special area formulas, etc.) to simplify calculations and sequentially determine the hypotenuse \(c\), perimeter \(L\), area \(S\), the two acute angles \(∠A\) and \(∠B\), the radius of the circumcircle \(R\), and the radius of the incircle \(r\). The specific methods are as follows:
Hint: In a right triangle, the side opposite the right angle is assumed to be the hypotenuse (i.e., \(c\) is the hypotenuse, corresponding to \(∠C=90^\circ\)). If the hypotenuse and one leg are known, the other leg can be derived by modifying the Pythagorean theorem. The core logic remains consistent.
1. How to calculate the hypotenuse \(c\)?
Directly apply the Pythagorean Theorem (the sum of the squares of the two legs equals the square of the hypotenuse), using the formula:
Take the positive square root to find the hypotenuse length:
2. How to calculate the perimeter \(L\)of the triangle?
The perimeter is the sum of the three sides. The formula is consistent with the previous section; substitute the lengths of the legs and hypotenuse:
3. How do you calculate the area \(S\)of a triangle?
Right triangles have a simplified formula (no trigonometric functions required), using the two legs as base and height:
Additional note: The general triangle formula for the area of a triangle between two sides and an angle (\(S = \frac{1}{2}ab\sin C\)) can also be used. Since \(∠C=90^\circ\) and \(\sin90^\circ=1\), the result is identical to the simplified formula. The simplified formula is more efficient.
4. How to calculate the two acute angles \(∠A\)and \(∠B\)?
The sum of the two acute angles in a right triangle is \(90^\circ\) (since the sum of interior angles is \(180^\circ\), minus the right angle \(90^\circ\)). We can first find one acute angle using the tangent function (opposite/adjacent), then use the sum of \(90^\circ\) to find the other. The steps are as follows:
Step 1: Find \( ∠A \) (opposite side \(a\), adjacent side \(b\)) using the tangent function formula:
Step 2: Find \( ∠B \)(using the property that two acute angles are complementary):
Hint: You can also calculate using the sine function (opposite side/hypotenuse) or the cosine function (adjacent side/hypotenuse), yielding the same result. The tangent function does not require finding the hypotenuse first, making the steps simpler.
5. How to calculate the radius \(R\)of a triangle’s circumcircle?
The circumcircle of a right triangle has a special property: its diameter equals the hypotenuse length. Thus, the radius formula simplifies to:
Verification: This matches the circumcircle radius formula for general triangles (\(R = \frac{a}{2\sin A}\)), since the opposite side of ∠A is \(a\), \(\sin A = \frac{a}{c}\), substituting yields \(R = \frac{a}{2 \times \frac{a}{c}} = \frac{c}{2}\), confirming the result.
6. How to calculate the radius \(r\)of a triangle’s inscribed circle?
For right triangles, a simplified formula exists (without needing to calculate area or semi-perimeter first):
Note: The general triangle formula for the inscribed circle radius (\(r = \frac{S}{p}\), where \(p = \frac{a + b + c}{2}\)) can also be used. Both methods yield the same result, but the specialized formula is faster.
Example Calculation
Given right triangle \(△ABC\) with \(∠C=90^\circ\), right angle side \(a=30\) (side \(BC\)), and right angle side \(b=40\) (side \(AC\)), the calculations proceed as follows:
1. Find hypotenuse \(c\):
Substitute into the Pythagorean theorem: \(c = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50\)
2. Find perimeter \(L\):
\(L = a + b + c = 30 + 40 + 50 = 120\)
3. Find the area \(S\):
Using the simplified formula for right triangles: \(S = \frac{1}{2} \times 30 \times 40 = 600\)
4. Find acute angles \(∠A\)and \(∠B\):
① Find \(∠A\): \(\tan A = \frac{a}{b} = \frac{30}{40} = 0.75\), thus \(∠A = \arctan(0.75) \approx 36.87^\circ\)
② Find \(∠B\): \(∠B = 90^\circ – 36.87^\circ = 53.13^\circ\)
5. Find the radius \(R\)of the circumcircle:
Substitute into the formula: \(R = \frac{c}{2} = \frac{50}{2} = 25\)
6. Find the radius \(r\)of the incircle:
Substitute into the formula: \(r = \frac{a + b – c}{2} = \frac{30 + 40 – 50}{2} = 10\)
Verification:
- The sum of the two acute angles \(36.87^\circ + 53.13^\circ = 90^\circ\) satisfies the property of complementary angles in a right triangle
- Verify using the general inscribed circle formula: Semicircumference \(p = \frac{30 + 40 + 50}{2} = 60\), \(r = \frac{S}{p} = \frac{600}{60} = 10\), consistent with the specialized formula result.