# Law of sines and cosines

In most of the practical applications, related to trigonometry, we need to calculate the angles and sides of a scalene triangle and not a right triangle. A scalene triangle is a triangle that has three unequal sides, each side having a different length. This implies that also all of the angles of the triangle are unequal to each other.

The most important laws related to scalene triangles are the law of sines and the law of cosines.

### Law of sines

Image: Law of sines for a scalene triangle

The scalene triangle above has:

• the sides a, b and c
• the angles αβ and γ
• the angle α is opposite to side aβ to b and γ to c
• all of the angles are acute angles

The circumscribed circle has the radius r.

The law of sines states that, in a scalene triangle, the ratio between one side and the sine of the opposite angle is equal with the diameter of the circumscribed circle.

$\bbox[#FFFF9D]{\frac{a}{\text{sin}(\alpha)}=\frac{b}{\text{sin}(\beta)}=\frac{c}{\text{sin}(\gamma)}=d} \tag{1}$

where d is the diameter of the circumscribed circle:

$d = 2 \cdot r \tag{2}$

The law of sines can also be interpreted as: in a scalene triangle, the ratio of two sides is equal with the ratio of the sine of the opposite angles.

$\begin{split} \frac{a}{b}=\frac{\text{sin}(\alpha)}{\text{sin}(\beta)}\\ \frac{a}{c}=\frac{\text{sin}(\alpha)}{\text{sin}(\gamma)}\\ \frac{b}{c}=\frac{\text{sin}(\beta)}{\text{sin}(\gamma)} \end{split}$

From equation (1) and (2) we can also extract the expressions of each side, function of the circumscribed circle’s radius and the sine of the opposite angle:

$\begin{split} a = 2 \cdot r \cdot \text{sin}(\alpha)\\ b = 2 \cdot r \cdot \text{sin}(\beta)\\ c = 2 \cdot r \cdot \text{sin}(\gamma) \end{split}$

The law of sines can be used to calculate the remaining sides of a triangle, when one side and two angles are known. This technique is also known as triangulation.

### Law of cosines

Image: Law of cosines for a scalene triangle

Pythagoras theorem is a particular case of the law of cosines.

The law of cosines states that, in a scalene triangle, the square of a side is equal with the sum of the square of each other side minus twice their product times the cosine of their angle.

$\begin{split}\bbox[#FFFF9D]{ a^2 = b^2 + c^2 – 2 \cdot b \cdot c \cdot \text{cos}(\alpha)\\ b^2 = a^2 + c^2 – 2 \cdot a \cdot c \cdot \text{cos}(\beta)\\ c^2 = b^2 + a^2 – 2 \cdot b \cdot a \cdot \text{cos}(\gamma)} \end{split}$

The law of cosines relates the length of each side of a triangle, function of the other sides and the angle between them.

If the angle is 90° (π/2), the law of cosines becomes Pythagoras’s theorem.

$\begin{split} \alpha = 90^{\circ} => a^2 = b^2 + c^2\\ \beta = 90^{\circ} => b^2 = a^2 + c^2\\ \gamma = 90^{\circ} => c^2 = b^2 + a^2 \end{split}$

In a scalene triangle, knowing the length of two sides and the angle between them, we can calculate the length of the third side. Also, knowing the length of each side, we can calculate all the angles:

$\begin{split} \alpha = \text{arccos} \left ( \frac{b^2 + c^2 – a^2}{2 \cdot b \cdot c} \right )\\ \beta =\text{arccos} \left ( \frac{a^2 + c^2 – b^2}{2 \cdot a \cdot c} \right )\\ \gamma =\text{arccos} \left ( \frac{b^2 + a^2 – c^2}{2 \cdot b \cdot a} \right ) \end{split}$

### Aircraft heading angle to compensate for wind

The law of sines and cosines has applicability in aircraft navigation. Calculating the necessary aircraft heading angle to compensate for the wind velocity and travel along a desired direction to a destination is a classic problem in aircraft navigation.

Image: Aircraft heading angle to compensate for wind

An aircraft is traveling with the speed (aircraft to wind) vaw = 150 m/s. Given the wind to ground speed vwg = 17 m/s and the wind angle γ = 15°, calculate the heading angle α [°] and the aircraft to ground speed vag [m/s]?

The three speed vectors are forming a scalene triangle. To find the heading angle α we can use the law of sines:

$\frac{v_{wg}}{\text{sin}(\alpha)}=\frac{v_{aw}}{\text{sin}(\gamma)}$

from which we can extract the expression of the heading angle α:

$\alpha = \text{arcsin} \left (\frac{v_{wg}}{v_{aw}} \cdot \text{sin}(\gamma) \right )$

Before replacing the numerical values, bear in mind that the arguments of the trigonometric functions need to be converted from degrees into radians.

The result of the arccos() function is in radians and needs to be converted back in degrees for an easier understanding.

$\alpha = \text{arcsin} \left (\frac{17}{150} \cdot \text{sin} \left ( 15^{\circ} \cdot \frac{\pi}{180^{\circ}} \right ) \right ) = 0.02934 \text{ rad} = 0.02934 \cdot \frac{180^{\circ}}{\pi} = 1.681^{\circ}$

We know that the sum of all the angles of any triangle is 180°. Therefore the angle β is:

$\beta = 180^{\circ} – \alpha – \gamma = 180^{\circ} – 17^{\circ} – 1.681^{\circ} = 161.32^{\circ}$

Knowing the aircraft speed, the wind speed and the angle between them, we can apply the law of cosines to calculate the aircraft ground speed:

$v_{ag} = \sqrt{v_{aw}^2+v_{wg}^2-2 \cdot v_{aw} \cdot v_{wg} \cdot \text{cos}(\beta)}$

Replacing the numerical values of the speeds and angle, gives:

$v_{ag} = \sqrt{150^2+17^2-2 \cdot 150 \cdot 17 \cdot \text{cos} \left ( 15 \cdot \frac{\pi}{180} \right ) } = 166.36 \text{ m/s}$

The aircraft will have a ground speed of 166.35 m/s and a heading angle of 1.681° to maintain the initial course (no wind case) towards destination.

Don’t forget to Like, Share and Subscribe!