Week 17: The Sine Law and The Cosine Law

Published by on June 10, 2018 | Leave a response
  • This week, trigonometry will be upgraded. The Sine Law and The Cosine Law are the general formulas that work based on the relationship between the sides and angles in triangles. If primary trigonometric ratios can only be applied to the right triangles, these Laws can be used for any triangles.

* The Sine Law:  \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} or \frac{\sin(A)}{a} = \frac{\sin(B)}{b} = \frac{\sin(C)}{c}

These two forms of The Sine Law can be used interchangeably depending on what the variable is. It’s easier to calculate when the variable is on the numerator. An easy way to remember the Sine Law is that the ratio of a side length and the corresponding sine of an angle is equal to that ratio of the other two. You can tell when to use this formula by looking at the given information: the triangle is given at a side and the corresponding angle.

*Importance: Remember to check to an ambiguous case. Since Sine is positive in the first and the second quadrant, there is a possibility to have two angles with the same reference angle as the answer.

  • Example: \triangle KMN has KM = 18cm, KN =15cm, \angle M =40. Find MN to the nearest tenth of a centimeter.

We have: \frac{\sin(N)}{n} = \frac{\sin(M)}{m}

\frac{\sin(N)}{18} = \frac{\sin{40}}{15}

\angle N = \sin^{-1} [\frac{18 (\sin {40})}{15}]

\angle N =50 or \angle N = 130

  • When \angle N = 50: \angle K = 180 – 50 – 40 -> \angle K =90
\frac{m}{\sin(M)} = \frac{k}{\sin(K)}

\frac{15}{\sin{40}} = \frac{k}{\sin{90}}

k= \frac{15(\sin {90})}{\sin {40}}

k=23

  • When \angle N = 130: \angle K = 180 – 130 – 40 -> \angle K =10

\frac{m}{\sin(M)} = \frac{k}{\sin(K)}

\frac{15}{\sin{40}} = \frac{k}{\sin{10}}

k= \frac{15(\sin{10})}{\sin{40}}

k= 4.1

* The Cosine Law: a^2 = b^2 + c^2 -2bc \cos{\alpha} or \cos{\alpha} = \frac{b^2 +c^2 - a^2}{2bc}

Depending on the asked information, these formulas can be put into use. The second one is the first formula after doing some algebra. The first form is used to find a side length, the second one is used to find the angle.

Example: In \triangle ABC, BC = 2AB, \angle B = 120, and AC=14cm. Determine the exact lengths of AB and BC.

We use x to represent AB -> BC = 2x

Using The Cosine Law, we have AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos{\beta}
14^2 = x^2 +(2x)^2 - 2x(2x)\cos{120}
196 = x^2 +4x^2 - 4x^2(\frac{-1}{2})
196 = 5x^2 + 2x^2
196 = 7x^2
28 = x^2
x = 2\sqrt{7}
2x = 4\sqrt{7}
So, AB= 2\sqrt{7} , BC = 4\sqrt{7}

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