TrigCalc.com

# Oblique Triangle Calculator

Area:
Perimiter:
Angle A:

Angle B:

Angle C:

Side A:

Side B:

Side C:

Area:
Perimiter:
Sin A:
Csc A:

Cos A:
Sec A:

Tan A:
Cot A:

Sin B:
Csc B:

Cos B:
Sec B:

Tan B:
Cot B:

Enter three known values (atleast one being a side)

area:

a:
b:
c:
A:
°
B:
°
C:
°

Only Two Values Required

An oblique triangle is a triangle without a 90 degree. All values can be calculated if either 1 side and any two other values are known. the law of sines and law of cosines are essential to the calculation process. These formulas are shown below:

\text{(law of sines)}

\frac{\text{sideA}}{\sin (\text{angleA})}=\frac{\text{sideB}}{\sin (\text{angleB})}=\frac{\text{sideC}}{\sin (\text{angleC})}

\text{law of cosines}

\text{sideA}^2=\text{sideB}^2+\text{sideC}^2-2\cdot \text{sideB}\cdot \text{sideC}\cdot \cos (\text{angleA})

\text{sideB}^2=\text{sideA}^2+\text{sideC}^2-2\cdot \text{sideA}\cdot \text{sideC}\cdot \cos (\text{angleB})

\text{sideC}^2=\text{sideA}^2+\text{sideB}^2-2\cdot \text{sideA}\cdot \text{sideB}\cdot \cos (\text{angleC})

Solving Triangles given two angles and one side:

If told to find the missing sides and angles of a triangle with angle A equaling 34 degrees, angle B equaling 58 degrees, and side a equaling a length of 16, you would begin solving the problem by determing with value to find first. Since two angle measures are already known, the third angle will be the simplest and quickest to calculate. To find the third angle (angle C) you can apply the law that all triangles add up to 180 degrees. this process is shown below:

\text{angleC}=180-\text{angleA}-\text{angleB}

\text{angleC}=180-(34^{\circ })-(58^{\circ })

\text{angleC}=88^{\circ }

\text{(in simplest form)}

To find the values of the other two sides, sides b and c, the law of sines must be used. The calculation process of finding these values is shown below:

\frac{\text{sideA}}{\sin (\text{angleA})}=\frac{\text{sideB}}{\sin (\text{angleB})}=\frac{\text{sideC}}{\sin (\text{angleC})}

\text{sideB}=\frac{(16)\sin ((58))}{\sin ((34))}

\text{sideB}=\frac{16\sin (58)}{\sin (34)}

\text{(in exact value form)}

24.26491726609088

\text{(in decimal approximation form)}

\text{sideC}=\frac{\text{sideA}\cdot \sin (\text{angleC})}{\sin (\text{angleA})}

\text{sideC}=\frac{(16)\sin ((88))}{\sin ((34))}

\text{sideC}=\frac{16\sin (88)}{\sin (34)}

\text{(in exact value form)}

28.5952363362601811

\text{(in decimal approximation form)}

Now that all sides and angles of the triangle have been calculated the area and perimiter of the triangle can be calculated. The calculation process for the area and perimiter is shown below:

\text{area}=\frac{1}{2}\cdot \text{sideA}\cdot \text{sideB}\cdot \sin (\text{angleC})

\text{area}=\frac{1}{2}(16)(\frac{16\sin (58)}{\sin (34)})\sin ((88))

\text{area}=\frac{128\sin (58)\sin (88)}{\sin (34)}

\text{(in simplest form)}

194.0010858728679993

\text{(in decimal form)}

Solving triangles given one angle and two sides (law of sines) If given one angle of a triangle and two sides, it is possible for two triangles to exist given the same dimensions. For example if told to find the missing sides and angles of a triangle given angle A is 19 degrees, side a is length 45, and side b length 44, you may begin by using the law of sines to find angle B. This is shown below:

\text{(law of sines)}

\frac{\sin (\text{angleA})}{\text{sideA}}=\frac{\sin (\text{angleB})}{\text{sideB}}=\frac{\sin (\text{angleC})}{\text{sideC}}

\text{angleB}=\sin ^{-1}(\frac{\text{sideB}\cdot \sin (\text{angleA})}{\text{sideA}})

\text{angleB}=\sin ^{-1}(\frac{(44)\sin ((19))}{(45)})

\text{angleB}=\sin ^{-1}(\frac{44\sin (19)}{45})/span>

\text{(in exact value form)}

\text{angleB}=18.5621602382506538^\circ

\text{(in exact decimal form)}

Since it is true that the sin of supplementary angles have equal values and the sin function was used to find angle B, you have to check if another triangle may exist with angle A measuring 19 degrees and and angle B measuring 18 degrees. Supplementary angles are angles who's sum equate to 180 degrees. If the sum of the supplement of angle B and angle A is less than 180 degrees, it is possible for two triangles to exist. To test this you can follow the process below:

180-\text{angleB}+\text{angleA}\textless180

180-(\sin ^{-1}(18.5621602382506538))+(19)\textless180

180.43783976174936>180

Since the sum of the supplement of angle B and angle A is greater than 180 degrees, it is impossible for another triangle to exist. You can continue to find the other sides and angles of the triangle. To find angle C you may follow the calculation process shown below:
\text{angleC}=180-\text{angleA}-\text{angleB}

\text{angleC}=180-(19^{\circ })-(\sin ^{-1}(18.5621602382506538)^{\circ })

\text{angleC}=142.4378397617493462^\circ

To find side C you can follow the calculation process shown below:
\text{sideC}=\frac{\text{sideA}\cdot \sin (\text{angleC})}{\sin (\text{angleA})}

\text{sideC}=\frac{(45)\sin ((-\sin ^{-1}(18.5621602382506538)+161))}{\sin ((19))}

\text{sideC}=84.2618657157949768^\circ

Finaly, the area of the triangle can be calculated using the calculation process shown below:
\text{area}=\frac{1}{2}\cdot \text{sideA}\cdot \text{sideB}\cdot \sin (\text{angleC})

\text{area}=\frac{1}{2}(45)(44)\sin ((-\sin ^{-1}(\frac{44\sin (19)}{45})+161))

\text{sideC}=84.2618657157949768^\circ