How to Use the Law of Sines to Solve Triangles
The law of sines is the relationship between angles and sides of all types of triangles such as acute, obtuse and right-angle triangles. It states the ratio of the length of sides of a triangle to sine of an angle opposite that side is similar for all the sides and angles in a given triangle.
To use the law of sines, you need to know either two sides of an angle of a triangle ( AAS or ASA) or two sides and an angle opposite one of them ( SSA). In this article, you will learn the law of sines, the law of sine definition, the law of sine formula, the law of sine problems, the law of sine proof, the law of sine applications and when to use law of sines.
Law of Sine Definition
The law of sine is defined as the ratio of the length of sides of a triangle to the sine of the opposite angle of a triangle. The law of sine is also known as Sine rule, Sine law, or Sine formula.
Law of sine is used to solve traingles.
\[\frac{a}{\text{Sin a}} = \frac{b}{\text{Sin b}} = \frac{c}{\text{Sin c}}\]
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a, b, and c are sides of the above triangle whereas A, B, and C are angles of above triangle.
( Sie a is opposite to angle A, side b is opposite to angle B, side c is opposite the angle C.)
(When we will divide side a by the sine of angle A, it will be equal to the side b by the sine of angle b and it will also be equivalent to side c by the sine of angle C).
Law of Sines Formula
The law of sine formula to find the unknown sides and angles of the triangle is given below
\[\frac{p}{\text{Sin P}} = \frac{q}{\text{Sin Q}} = \frac{r}{\text{Sin R}}\]
p :q : r = Sin P : Sin Q: Sin R
\[\frac{p}{q} = \frac{\text{Sin P}}{\text{Sin Q}}\]
\[\frac{q}{r} = \frac{\text{Sin Q}}{\text{Sin R}}\]
The above law of sine formula represents that if we will divide side a by the sine of angle A, it will be equal to the side b by the sine of angle b and it will also be equivalent to side c by the sine of angle C).
Here, Sin A represents a number whereas a is the length of a triangle.
Law of Sine Proof
For the law of sine proof, we need to have right-angle triangles as trigonometry functions are mostly defined in terms of right-angle triangles only.
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Given: △ ABC
Construction: Draw a perpendicular line i.e. CD⊥ AB, then CD = height of the triangle. “H” divides the ABC into two right-angled triangles.
To proof : \[\frac{a}{b} = \frac{\text{Sin a}}{\text{Sin b}}\]
Proof: In △ CDA
Sin A = \[\frac{h}{b}\]
And in △ CDB
Sin B = \[\frac{h}{a}\]
Hence, \[\frac{\text{Sin A}}{\text{Sin B}} = \frac{(\frac{h}{b})}{(\frac{h}{a})} = \frac{a}{b}\]
Hence, proved
Similarly, we can prove \[\frac{Sin B}{Sin C} = \frac{b}{c}\] and other corresponding pairs of angles and their opposite sides.
Law of Sine - Ambiguous Case
If in a triangle any two sides and their opposite angles are known, then there are three possibilities in the law of sine- ambiguous case. M these three possibilities are:
That there is no such triangle
There are two distinct triangle
There is exactly one such triangle
When to Use the Law of Sines?
The law of sine is used to find the angles of an ordinary triangle. In two sides and the enclosed angles are given, it can be simultaneously used to find the third side and other two angles.
If two sides and angles opposite one of them are known then the other can be calculated.
The third angle is then calculated by the fact that the sum of the interior angles of a triangle is equal to 180 degrees.
We use the following law of sine formula to calculate the unknown sides and angles of a triangle.
\[\frac{a}{\text{Sin a}} = \frac{b}{\text{Sin b}} = \frac{c}{\text{Sin c}}\]
Law of sine problem
Here, you can see the law of sine problem with the solution
Calculate △PQR in which ∠ P = 116°, p = 8.3 cm and q - 5.4 cm.
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Solution: \[\frac{Sin 116^{\circ}}{8.3}\] = \[\frac{SinQ}{5.4}\]
Sin Q = \[5.4\frac{Sin 116^{\circ}}{8.3}\] = 0.5848
Q = = Sin-¹ 0.5848
35.8° or 144.2°
Q cannot be an obtuse angle as the sum of the interior angle of a triangle will exceed
180°. The only value for Q will be considered is 35.8°.
∠Q = 35.8° , ∠ R = 180° - 116° - 35.8° = 28.2°
\[\frac{8.3}{Sin 116^{\circ}}\] = \[\frac{r}{Sin 28.2^{\circ}}\]
r = \[8.3\frac{Sin 28.2^{\circ}}{Sin 116^{\circ}}\] = 4.36
Hence, in the above triangle ∠Q = 35.8°, ∠ R = 28.2° and side r = 4.36 cm
Quiz Time
1. If, c= 6cm, b = 9 cm and a = 14 cm. Find A
133
137
188
121
2. In triangle ABC, if ∠A = 98° and ∠B = 12°. If side a is 84 units long approximately then find the length of side b.
18 units
17 units
12 units
86 units
Solved Examples
Here, you can see some of the law of sine example problem
1. If ∠B = 21°, ∠c = 46 and side AB = 9 cm in a triangle then solve the triangle.
Solution:
Given: Two angles and a side
No, let’s use law of sine to solve the triangle
As we know sum of the interior sides of a triangle is 180°
Accordingly, ∠A = 113°
As AB = c= 9cm
Using the law of sine rule:
\[\frac{a}{Sin 113^{\circ}} = \frac{b}{Sin 21^{\circ}} = \frac{9}{Sin 46^{\circ}}\]
\[\frac{b}{Sin 21^{\circ}} = \frac{9}{Sin 46^{\circ}}\]
b = Sin 21° × \[\frac{9}{Sin 46^{\circ}}\]
= 4.484 cm
A = sin 113° × \[\frac{9}{Sin 46^{\circ}}\]
= 11.517 cm.
2. In PQR, Sin p = 1/3 and Sin q = ¼, Find the ratio of side p/side q
Solution:
Using the law of sine, we get
\[\frac{p}{\text{Sin P}} = \frac{q}{\text{Sin Q}} = \frac{p}{\frac{1}{3}} = \frac{q}{\frac{1}{4}}\]
\[\frac{1}{3p}\] = \[\frac{1}{4q}\]
\[\frac{4}{1}\times \frac{1}{3q} = p\]
\[\frac{4}{3} = \frac{p}{q}\]
Hence, the ratio of the side p to side q is 4 to 3.
FAQs on Law of Sines: Definition, Formula & Applications
1. What is the Law of Sines?
The Law of Sines, also known as the Sine Rule, is a fundamental principle in trigonometry that establishes a relationship between the lengths of the sides of any triangle and the sines of their opposite angles. It is particularly useful for solving oblique triangles (triangles that do not have a right angle).
2. What is the formula for the Law of Sines?
For any triangle with sides of length a, b, and c, and opposite angles A, B, and C respectively, the Law of Sines formula is expressed as a ratio:
a/sin(A) = b/sin(B) = c/sin(C)
This constant ratio is also equal to 2R, where 'R' is the radius of the triangle's circumcircle.
3. Under what conditions should the Law of Sines be used to solve a triangle?
The Law of Sines is the appropriate tool to find unknown sides or angles when you have the following information:
- Angle-Angle-Side (AAS): You know two angles and one non-included side.
- Angle-Side-Angle (ASA): You know two angles and the side included between them.
- Side-Side-Angle (SSA): You know two sides and a non-included angle. This is also known as the ambiguous case and requires careful analysis.
4. What are some real-world examples of the Law of Sines in action?
The Law of Sines has several practical applications in various fields. For example:
- Navigation and Aviation: It is used to calculate distances and bearings, such as determining a ship's distance from two different lighthouses or an aircraft's location.
- Surveying: Land surveyors use it to measure the distance between two inaccessible points by measuring angles from a known baseline.
- Astronomy: It helps in calculating the distances between celestial bodies.
- Engineering: In structural engineering, it helps in calculating the forces acting on the members of a triangular truss.
5. How is the Law of Sines different from the Law of Cosines, and how do you decide which to use?
The choice between the Law of Sines and the Law of Cosines depends on the given information about the triangle.
Use the Law of Sines when you know a side and its opposite angle (e.g., AAS, ASA, or SSA cases).
Use the Law of Cosines when you know two sides and the angle between them (SAS case) or when you know all three sides (SSS case). The Law of Cosines is essential when no angle-opposite-side pair is known.
6. What is the 'ambiguous case' (SSA) in the Law of Sines and why does it occur?
The ambiguous case occurs when you are given two sides and a non-included angle (SSA). It is called 'ambiguous' because this set of information can result in zero, one, or two possible triangles. This ambiguity arises because the sine function is positive in both the first and second quadrants. Therefore, when solving for an unknown angle using sin⁻¹, there can be two possible values (e.g., θ and 180°-θ) that need to be checked for validity within the triangle.
7. Can the Law of Sines be used for a right-angled triangle? Explain why or why not.
Yes, the Law of Sines can be applied to a right-angled triangle, but it is generally an overly complicated method. The basic trigonometric ratios (SOH-CAH-TOA) are far more direct and efficient. For instance, if angle C is 90°, then sin(C) = 1. The Law of Sines, a/sin(A) = c/sin(C), simplifies to a/sin(A) = c/1, which rearranges to sin(A) = a/c. This is the standard definition of sine (opposite/hypotenuse), demonstrating that the Law of Sines is consistent but redundant for right-angled triangles.
