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From Wikipedia

Law of sines

In trigonometry, the law of sines (also known as the sine law, sine formula, or sine rule) is an equation relating the lengths of the sides of an arbitrary triangle to the sines of its angles. According to the law,

\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C},

where a, b, and c are the lengths of the sides of a triangle, and A, B, and C are the opposite angles (see the figure to the right). Sometimes the law is stated using the reciprocal of this equation:

\frac{\sin A}{a} \,=\, \frac{\sin B}{b} \,=\, \frac{\sin C}{c}.

The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the formula gives two possible values for the enclosed angle, leading to an ambiguous case.

The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in a general triangle, the other being the law of cosines.


The following are examples of how to solve a problem using the law of sines:

Given: side a = 20, side c = 24, and angle C = 40°

Using the law of sines, we conclude that

\frac{\sin A}{20} = \frac{\sin 40^\circ}{24}.
A = \arcsin\left( \frac{20\sin 40^\circ}{24} \right) \cong 32.39^\circ.

Or another example of how to solve a problem using the law of sines:

If two sides of the triangle are equal to R and the length of the third side, the chord, is given as 100 feet and the angle C opposite the chord is given in degrees, then

\angle A = \angle B = \frac{180-C}{2}= 90-\frac{C}{2}


{R \over \sin A}={\mbox{chord} \over \sin C}\text{ or }{R \over \sin B}={\mbox{chord} \over \sin C}\,
{\mbox{chord} \,\sin A \over \sin C} = R\text{ or }{\mbox{chord} \,\sin B \over \sin C} = R.

Numeric problems

Like the law of cosines, although the law of sines is mathematically true, it has problems for numeric use. Much precision may be lost if an arcsine is computed when the sine of an angle is close to one.

Some applications

  • The sine law can be used to prove the angle sum identity for sine when α and β are each between 0 and 90 degrees.
To prove this, make an arbitrary triangle with sides a, b, and c with corresponding arbitrary angles A, B and C. Draw a perpendicular to c from angle C. This will split the angle C into two different angles, α and β, that are less than 90 degrees, where we choose to have α to be on the same side as A and β be on the same side as B. Use the sine law identity that relates side c and side a. Solve this equation for the sine of C. Notice that the perpendicular makes two right angles triangles, also note that sin(A) = cos(α), sin(B) = cos(β) and that c = a sin(β) + b sin(α). After making these substitutions you should have sin(C) =sin(α + Î²) = sin(β)cos(α) + (b/a)sin(α)cos(α). Now apply the sine law identity that relates sides b and a and make the substitutions noted before. Now substitute this expression for (b/a) into the original equation for sin(α + Î²) and you will have the angle sum identity for α and β in terms of sine.
The only thing that was used in the proof that was not a definition was the sine law. Thus the sine law is equivalent to the angle sum identity when the angles sum is between 0 and 180 degrees and when each individual angle is between 0 and 90 degrees.

The ambiguous case

When using the law of sines to solve triangles, there exists an ambiguous case where two separate triangles can be constructed (i.e., there are two different possible solutions to the triangle).

Given a general triangle ABC, the following conditions would need to be fulfilled for the case to be ambiguous:

  • The only information known about the triangle is the angle A and the sides a and b, where the angle A is not the included angle of the two sides (in the above image, the angle C is the included angle).
  • The angle A is acute (i.e., A < 90°).
  • The side a is shorter than the side b (i.e., a< b).
  • The side a is longer than the altitude of a right angled triangle with angle A and hypotenuse b (i.e., a> b sin A).

Given all of the above premises are true, the angle B may be acute or obtuse; meaning, one of the following is true:

B = \arcsin {b \sin A \over a}


B= 180^\circ - \arcsin {b \sin A \over a}

Relation to the circumcircle

In the identity

\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C},

the common value of the three fractions is actually the diameter of the triangle's circumcircle. It can be shown that this quantity is equal to


\frac{abc} {2S} & {} = \frac{abc} {2\sqrt{s(s-a)(s-b)(s-c)}} \\[6pt] & {} = \frac {2abc} {\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4) }}, \end{align}

where S is the area of the triangle and s is the semiperimeter

s = \frac{a+b+c} {2}.

The second equality above is essentially Heron's formula.

Spherical case

In the spherical case, the formula is:

\frac{\sin A}{\sin \alpha} = \frac{\sin B}{\sin \beta} = \frac{\sin C}{\sin \gamma}.

Here, α, β, and γ are the angles at the center of the sphere subtended by the three arcs of the spherical surface triangle a,b, and c, respectively. A, B, and C are the surface angles opposite their respective arcs.

See also From Yahoo Answers

Question:A car travels along a straight road, heading east for 1 h, then traveling for 30 min on another road that leads northeast. If the car has maintained a constant speed of 62 mi/h, how far is it from its starting position? Round your answer to 2 decimal places. how do you do this?

Answers:First off we must find how far we have driven on each road. This is fairly easy, since they give you the speed (62mph) and the time traveled is 1 hour and then 30 min. So, the car headed east for 62 miles, and then headed north east (45 degrees) for 31 miles. Now, the angle involved in the equation, will be 180 - 45 = 135 degrees. Now use the formula (law of cosines): ........_________________ c = a^2 + b^2 - 2ab(cos y) ........_________________________ c = 62^2 + 31^2 - 2(62)(31)(cos135) ........______________________ c = 3844 + 961 - 3844(-.7071) ........____________________ c = 3844 + 961 + 2718.0924 ........_________ c = 7523.0924 c = 86.74 miles :)

Question:I need a little help refreshing my memory. I'm redoing this problem that I did wrong on vectors and if sine, cosine, and tangent can only be used to find lengths of a side of only right triangles, that would explain so much why I did it wrong... As for law of sines and law of cosines, can they be used for any triangles like scalene, isosceles, etc.? I have the worst memory in math and I wanna know so I can add to my old math notes compiled over 3 years of high school math and hopefully I won't forget.

Answers:sin x = a/c, cosx = b/c .... can be used only for right triangles law of sines and cosines can be used in any triangle

Question:Describe each of the following properties of the graph of the Sine Function, f(theta) = sin(theta) and relate the property to the unit circle definition of sine. * Amplitude * Period * Domain * Range * x-intercepts I wrote: The Amplitude of a sine graph is the absolute value of the value a in y=d+a(b(x-c), it relates to the radius of the unit circle. The period can be found by dividing 2pi by b in the equation above. The domain of a sine graph is negative infinity to infinity and the range is from -1 to 1. The x-intercepts are found by dividing c+period by half, then c+(c+period) by half and period+(c+period). The height of the sin curve is equal to the y value of the unit circle My teacher responded: Good attempt but some of this is unclear. As the problem asked for the relationship to the unit circle you should try to keep that in mind when writing. Avoid redefining terms unless defining them specifically with their relationship to the unit circle. Can someone please help me clean up my answer, I am unsure of what to write?

Answers:I will use t instead of theta. The question asks about f(t) = sin(t), not about some more general sine function, so I would answer as follows: * Amplitude: 1, because largest y-coordinate of unit circle is 1 * Period: 2pi because circumference of unit circle is 2pi and t is measured as arc-length around the circle. * Domain: All real numbers, because t can go as many times around the circle, both clockwise (positive) and counterclockwise (negative) as you like. * Range: [-1, 1] because the y-coordinate on the unit circle ranges between -1 and 1 inclusive * x-intercepts: values of t, where y=0, namely 0 plus any integer multiple of 2pi, and pi + any integer multiple of 2pi, or, more succinctly, k pi where k is any integer.

Question:I have a math assignment, and I need to use those laws. However I can't find my notes or anything in the textbook about it. Could you please explain these laws to me please? Thank you in advance!

Answers:here is the link. The link shows three eqns. so juz memorise one will do. the law is used when only 1 angleand 2 length are given. So using this law to find the other length.

From Youtube

Trig: Applying Law of Cosines, Law of Sines :www.mindbites.com This 77 minute trigonometry lesson covers oblique triangles and will help you understand the difference between a right triangle and an oblique triangle, recognize acute and oblique triangles and how to work with: - the Law of Cosines for the ASA (angle side angle) and SSS (side side side) situations - the Law of Sines for the AAS (angle angle side) ASA (angle side angle) and SSA (side side angle) situations - the Law of Sines plus the ambiguous case Thislesson contains explanations of the concepts and 15 example questions with step by step solutions plus 5 interactive review questions with solutions. Lessons that will help you with the fundamentals of this lesson include: - 300 Basic Trigonometry Part I (www.mindbites.com - 305 Solving Word Problems Using Sine, Cosine and Tangent (www.mindbites.com

Trig: Word Problems - Sine, Cosine & Tangent :www.mindbites.com This 78 minute trigonometry lesson is an opportunity to see how the skills learned in Lesson 300 Basic Trigonometry Part I are used to solving word problems using the 3 primary trig ratios, sin, cos and tan. You find unknown sides and angles in right angle triangles to solve word problems that involve: - a kite on the end of a string, diagonal of a rectangle, support wires, shadows, ladders, trees, buildings, stairways and more - angle of elevation and angle of depression - rise & run Sample problem: An airplane is flying over a forest at an altitude of 1600 ft. when the pilot sights a fire. He measures the angle of depression and finds it to be 46 degrees. How far is the fire, to the nearest tenth of a ft. from a point on land directly below the airplane? This lesson contains explanations of the concepts and 17 example questions with step by step solutions plus 3 interactive review questions with solutions. The following lesson will help you with the fundamentals of this lesson: - 300 Basic Trigonometry Part I (www.mindbites.com