Sine and Cosine Rule

sine and cosine

Sine Rule

 

Trigonometry is also usable in non-right angle triangles, and  Sine Rule is an important law relating to sides and lengths for any triangle. This rule is used to find:

  1. side, given one side and two angles
  2. an angle, given two sides and one angle

\cfrac { a }{ \sin { A }  } =\cfrac { b }{ \sin { B }  } =\cfrac { c }{ \sin { C }  }

The idea here is to compare two pairs of sides and their opposite angles, as the ratio of the sides and the sine of their opposite angles have the same ratio value.

 

Examples:

 

1. Find the length of side p.

Show Answer

\cfrac { p }{ \sin { 32 }  } =\cfrac { 21 }{ \sin { 95 }  } \\ \\ p=\cfrac { 21 }{ \sin { 95 }  } \times \sin { 32 } \\ \\ p=11.17\quad units

 

2. Find the value of angle b.

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\cfrac { 3.6 }{ \sin { b }  } =\cfrac { 5.1 }{ \sin { 100 }  } \\ \\ \cfrac { \sin { b }  }{ 3.6 } =\cfrac { \sin { 100 }  }{ 5.1 } \qquad \longleftarrow \quad reciprocal\\ \\ \sin { b } =3.6\times \cfrac { \sin { 100 }  }{ 5.1 } \\ \\ \sin { b } =0.695\\ \\ b=\sin ^{ -1 }{ 0.695 } \\ \\ b=44^{ \circ  }

 

Cosine Rule

 

Another important rule is the cosine rule.

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

Use cosine rule to find:

  1. side, given two sides and one angle
  2. an angle, given three sides

 

Examples:

 

1. From the triangle below, find the length of the side QR.

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{ QR }^{ 2 }={ 13 }^{ 2 }+{ 4 }^{ 2 }-2\left( 13 \right) \left( 4 \right) \cos { 70^{ \circ  } } \\ \\ { QR }^{ 2 }=169+16-104\cos { 70^{ \circ  } } \\ \\ { QR }^{ 2 }=149.43\\ \\ QR=\sqrt { 149.43 } =12.22cm

 

2. Is it possible for a triangle to have the following dimensions (lengths and angles)?

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This question is inviting a detective work. Let's check whether the angle is really 110°.

{ 73 }^{ 2 }={ 50 }^{ 2 }+{ 55 }^{ 2 }-2\left( 50 \right) \left( 55 \right) \cos { \theta  } \\ \\ 5329=2500+3025-5500\cos { \theta  } \\ \\ 5329=5525-5500\cos { \theta  } \\ \\ -196=-5500\cos { \theta  } \\ \\ \theta =\cos ^{ -1 }{ \cfrac { -196 }{ -5500 }  } \\ \\ \theta =87.96^{ \circ  }

Therefore, the triangle is not possible, because the angle opposite of the side of length 73m must be 87.96°, not 110°.

 

Area of a Triangle (Sine)

 

To find the area of a triangle without a given height, trigonometry provides a solution using sine. Using the triangle below a formula could be conjured:

Area=\cfrac { 1 }{ 2 } ab\sin { C }

 

Example:

Find the area of the triangle below (lengths are in cm).

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Area=\cfrac { 1 }{ 2 } \left( 10.8 \right) \left( 9.8 \right) \sin { 54.3^{ \circ  } } \\ \\ Area=42.98{ cm }^{ 2 }