Triangle is a shape in the world of geometry. The word triangle can be broken into two parts. The first part is the tri and it means three and the second part is the angle. Hence, the meaning of triangle is a geometry of three lines which are making three angles. However, there are a few types of a triangle that differentiate from other triangles. The variation you will find will be in angles and lengths. Below are the types of triangles:

  1. Equilateral Triangle
  2. Isosceles Triangle
  3. Oblique Triangle

An oblique triangle does not have a right angle and can also be classified as an acute triangle or an obtuse triangle. The specialty of an oblique triangle is that it has all different angles and different lengths.

To solve oblique triangles, use the laws of sine and cosine. There are four different potential scenarios:

1. Solve a Triangle Knowing: One Side and Two Adjacent Angles.

 

A = { 180 }^{ \circ } - B - C

\frac { a }{ \sin { A } } = \frac { b }{ \sin { B } } \qquad b = a . \frac { \sin { B } }{ \sin { A } }

\frac { a }{ \sin { A } } = \frac { c }{ \sin { C } } \qquad c = a . \frac { \sin { C } }{ \sin { A } }

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Example

Solve the oblique triangle with the following data: a = 6 m, B = { 45 }^{ \circ } and C = { 105 }^{ \circ }.

A = { 180 }^{ \circ } - { 45 }^{ \circ } - { 105 }^{ \circ } = { 30 }^{ \circ }

\frac { 6 }{ \sin { { 30 }^{ \circ } } } = \frac { b }{ \sin { { 45 }^{ \circ } } } \qquad b = 6 \times \frac { \sin { { 45 }^{ \circ } } }{ \sin { { 30 }^{ \circ } } }

b = 6 \times \frac { \frac { \sqrt { 2 } }{ 2 } }{ \frac { 1 }{ 2 } } = 6 \sqrt { 2 } m

 

\frac { 6 }{ \sin { { 30 }^{ \circ } } } = \frac { c }{ \sin { { 105 }^{ \circ } } } \qquad c = 6 \times \frac { \sin { { 105 }^{ \circ } } }{ \sin { { 30 }^{ \circ } } }

c = 11.6 m

 

2. Solve a Triangle Knowing Two Sides and an Included Angle.

c = \sqrt { { a }^{ 2 } + { b }^{ 2 } - 2 . a . b \cos { C }}

\frac { b }{ \sin { B } } = \frac { c }{ \sin { C } }  \qquad \sin { B } = \frac { c . \sin { B } }{ b }

A = { 180 }^{ \circ } - B - C

Example

Solve the oblique triangle with the following data: a = 10 m, b = 7 m and C = { 30 }^{ \circ }°.

c = \sqrt { { 10 }^{ 2 } + { 7 }^{ 2 } - 2 . 10 . 7 \cos { { 30 }^{ \circ } } } = 5.27 m

\frac { 7 }{ \sin { B } } = \frac { 5.27 }{ \sin { { 30 }^{ \circ } } } \qquad \sin { B } = 0.664 \Rightarrow \left\{\begin{matrix} B = { 41.60 }^{ \circ }  \\ B = { 138.40 }^{ \circ } \end{matrix}\right

B = { 41.60 }^{ \circ } because a > b and that is why, obtuse angle is A.

A = { 180 }^{ \circ } - { 30 }^{ \circ } - { 41.60 }^{ \circ } = { 108.4 }^{ \circ }

3. Solve a Triangle Knowing Two Sides and the Opposite Angle.

\frac { a }{ \sin { A } } = \frac { b }{ \sin { B } }, If:

\sin { B } > 1 will result in no solution.

\sin { B } = 1 will result in one solution.

\sin { B } < 1 will result in one or two solutions.

1. sin B > 1. No solution

Solve the triangle with the following data: A = { 30 }^{ \circ }, a = 3 m and b = 8 m.

\frac { 3 }{ \sin { { 30 }^{ \circ } }} = \frac { 8 }{ \sin { B } } \qquad \sin { B } = \frac { 4 }{ 3 } > 1

Since the sine of an angle can never be greater than 1, the problem has no solution. The drawing above shows the impossibility of the situation.

 

2. sin B = 1. One Solution: Right Triangle

Solve the triangle with the following data: A = { 30 }^{ \circ }, a = 3 m, and b = 6 m.

\frac { 3 }{ \sin { { 30 }^{ \circ } }} = \frac { 6 }{ \sin { B } } \qquad \sin { B } = \frac { 3 }{ 3 } = 1 \qquad B = { 90 }^{ \circ }

C = { 90 }^{ \circ } - { 30 }^{ \circ } = { 60 }^{ \circ }

 

a = 6 . \sin { { 30 }^{ \circ } } = 3m

c = 6 . \cos { { 30 }^{ \circ } } = 3 \sqrt { 3 } m

 

3. sin B < 1. One or Two Solutions

Solve the triangle with the following data: A = { 60 }^{ \circ }, a = 8 m, and b = 4 m.

\frac { 8 }{ \sin { { 60 }^{ \circ } } } = \frac { 3 }{ \sin { B } } \qquad \sin { B } = \frac { \sqrt { 3 } }{ 4 } \Rightarrow \left\{\begin{matrix} B = { 25.65 }^{ \circ } \\ B= { 180 }^{ \circ } - { 25.65 }^{ \circ } = { 154.35 }^{ \circ } \end{matrix}\right

As a > b there is one solution which is B = { 25.65 }^{ \circ }

C = { 180 }^{ \circ } - ({ 60 }^{ \circ } + { 25.65 }^{ \circ }) = { 94.35 }^{ \circ }

\frac { 8 }{ \sin { { 60 }^{ \circ } }} = \frac { c }{ \sin { { 94.35 }^{ \circ } } } \qquad c = 9.21m

 

Solve the triangle with the following data: A = { 30 }^{ \circ }, a = 3 m, and b = 4 m.

\frac { 3 }{ \sin { { 30 }^{ \circ } } } = \frac { 4 }{ \sin { B } } \qquad \sin { B } = \frac { 2 }{ 3 } \Rightarrow \left\{\begin{matrix} { B }_{ 1 } = { 41.81 }^{ \circ }  \\ { B }_{ 2 } = { 180 }^{ \circ } - { 41.81 }^{ \circ } = { 138.19 }^{ \circ } \end{matrix}\right

As a < b therefore, there are two solutions:

{ C }_{ 1 } = { 180 }^{ \circ } - ({ 30 }^{ \circ } + { 41.81 }^{ \circ }) = { 108.19 }^{ \circ }

\frac { 3 }{ \sin { { 30 }^{ \circ } }} = \frac { c }{ \sin { 108.19 }} \qquad c = 5.7m

{ C }_{ 2 } = { 180 }^{ \circ } - ({ 30 }^{ \circ } + { 138.19 }^{ \circ }) = { 11.81 }^{ \circ }

\frac { 3 }{ \sin { { 30 }^{ \circ } } } = \frac { c }{ \sin { { 11.81 }^{ \circ } } } \qquad c = 1.227m

 

4. Solve a Triangle Knowing Two Sides and the Opposite Angle.

\cos { A } = \frac { { b }^{ 2 } + { c }^{ 2 } - { a }^{ 2 } }{ 2 . b . c }

\cos { B } = \frac { { a }^{ 2 } + { c }^{ 2 } - { b }^{ 2 } }{ 2 . a . c }

C = { 180 }^{ \circ } - A - B

Solve the triangle with the following data: a = 15 m, b = 22 m, and c = 17 m.

\cos { A } = \frac { 484 + 289 - 225 }{ 748 } = 0.7326 \qquad A = { 42.89 }^{ \circ }

\cos { B } = \frac { 289 + 225 - 484 }{ 510 } = 0.0588 \qquad A = { 86.62 }^{ \circ }

{ 180 }^{ \circ } - { 42.89 }^{ \circ } - { 86.62 }^{ \circ } = { 50.49 }^{ \circ }

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.