正弦定律

Illustrated · 1 min

The law of sines of a triangle is a formula relating the angles of a triangle to their corresponding side lengths. It states the following.

First, let ABCABC be a triangle.

Let a=BCa=BC, b=ACb=AC, and c=ABc=AB. Then, according to the law of sines: sin⁡∠Aa=sin⁡∠Bb=sin⁡∠Cc.\frac {\sin \angle A}{a} = \frac {\sin \angle B}{b} = \frac {\sin \angle C}{c}.

Proof. The area of triangle ABCABC can be written in three different ways: [△ABC]=12bcsin⁡∠A=12acsin⁡∠B=12absin⁡∠C,[\triangle ABC] = \frac {1}{2} bc \sin \angle A = \frac {1}{2} ac \sin \angle B = \frac {1}{2} ab \sin \angle C, giving us the result. ■\blacksquare

Put ideas together

Contest problems often need this idea together with one or two others. That is where understanding goes deeper than memory. These are real problems from the AMC 12 and AIME, the national math competitions for middle and high school students, each with a worked solution on LIVE by Po-Shen Loh.

Illustrated · 2 min

Review of The Law of Sines

The Law of Sines, also known as the Sine Rule, is useful for solving angles of a triangle. The law is as follows: Asin⁡(a)=Bsin⁡(b)=Csin⁡(c)\frac{A}{\sin(a)} = \frac{B}{\sin(b)} = \frac{C}{\sin(c)} Where a,ba,b and cc are angles, and A,BA,B and CC are sides.

What this law states is that when we divide side aa by the sine of angle AA it is equal to side bb by the sine of angle BB, and also equal to side cc divided by the sine of angle CC

Example of The Law of Sines

A triangle with angles labeled a, b = 43.2, c = 87.9 and sides labeled A, 7, C

Creative Commons for Commercial Reuse by Sarah Walker at expii.com

To start out this problem, let's write out the law of sines. Asin⁡(a)=Bsin⁡(b)=Csin⁡(c)\frac{A}{\sin(a)} = \frac{B}{\sin(b)} = \frac{C}{\sin(c)} We can see from above that the a,Aa, A and CC are what we will be solving for, so that means we do have our b,Bb, B and cc terms.

So, plugging these into the law of sines, we will have: Asin⁡(a)=7sin⁡(43.2∘)=Csin⁡(87.9∘)\frac{A}{\sin(a)} = \frac{7}{\sin(43.2^{\circ})} = \frac{C}{\sin(87.9^{\circ})}

We notice that we have no information for neither AA nor aa. However, we can easily find aa given simple geometry! We know that the angles of a triangle add up to 180∘180^{\circ}. If you need a refresher on the addition of angles in a triangle, you can click here! So, we can do a=180−b−ca = 180 - b - c a=180−43.2−87.9a = 180 - 43.2 - 87.9 This will give us an aa of 48.9∘48.9^{\circ}! Let's update our law of sines: Asin⁡(48.9∘)=7sin⁡(43.2∘)=Csin⁡(87.9∘)\frac{A}{\sin(48.9^{\circ})} = \frac{7}{\sin(43.2^{\circ})} = \frac{C}{\sin(87.9^{\circ})}

Now, let's compare our B/bB/b and C/cC/c terms. This will look like: 7sin⁡(43.2∘)=Csin⁡(87.9∘)\frac{7}{\sin(43.2^{\circ})} = \frac{C}{\sin(87.9^{\circ})}

Let's multiply both sides by sin⁡(43.2∘)sin⁡(87.9∘)\sin(43.2^{\circ}) \sin(87.9^{\circ}): 7sin⁡(43.2∘)×sin⁡(43.2∘)sin⁡(87.9∘)=Csin⁡(87.9∘)×sin⁡(43.2∘)sin⁡(87.9∘)\frac{7}{\sin(43.2^{\circ})} \times \textcolor{red}{\sin(43.2^{\circ}) \sin(87.9^{\circ})} = \frac{C}{\sin(87.9^{\circ})} \times \textcolor{red}{\sin(43.2^{\circ}) \sin(87.9^{\circ})}

This will give us: 7sin⁡(87.9∘)=Csin⁡(43.2∘)7 \sin(87.9^{\circ}) = C \sin(43.2^{\circ})

Now, if we divide both sides by sin⁡(43.2∘)\sin(43.2^{\circ}), we will find our CC term!

7sin⁡(87.9∘)sin⁡(43.2∘)=Csin⁡(43.2∘)sin⁡(43.2∘)\frac{7 \sin(87.9^{\circ})}{\textcolor{red}{\sin(43.2^{\circ})}} =\frac{C \sin(43.2^{\circ})}{\textcolor{red}{\sin(43.2^{\circ})}} C=7sin⁡(87.9∘)sin⁡(43.2∘)C = \frac{7 \sin(87.9^{\circ})}{\sin(43.2^{\circ})} C=10.2C = 10.2

Great! Now we only have AA to find. Since we were given the B/bB/b terms in the problem, we know that those are for certain the answer. Let's use them again to find AA. This will look like:

Asin⁡(48.9∘)=7sin⁡(43.2∘)\frac{A}{\sin(48.9^{\circ})} = \frac{7}{\sin(43.2^{\circ})}

Let's multiply both sides by sin⁡(48.9∘)sin⁡(43.2∘)\sin(48.9^{\circ}) \sin(43.2^{\circ}):

Asin⁡(48.9∘)×sin⁡(48.9∘)sin⁡(43.2∘)=7sin⁡(43.2∘)×sin⁡(48.9∘)sin⁡(43.2∘)\frac{A}{\sin(48.9^{\circ})} \times \textcolor{red}{\sin(48.9^{\circ}) \sin(43.2^{\circ})} = \frac{7}{\sin(43.2^{\circ})} \times \textcolor{red}{\sin(48.9^{\circ}) \sin(43.2^{\circ})}

This will give us: Asin⁡(43.2∘)=7sin⁡(48.9∘)A \sin(43.2^{\circ}) = 7 \sin(48.9^{\circ})

Now, if we divide both sides by sin⁡(43.2∘)\sin(43.2^{\circ}), we will find our AA term!

Asin⁡(43.2∘)sin⁡(43.2∘)=7sin⁡(48.9∘)sin⁡(43.2∘)\frac{A \sin(43.2^{\circ})}{\textcolor{red}{\sin(43.2^{\circ})}} =\frac{7 \sin(48.9^{\circ})}{\textcolor{red}{\sin(43.2^{\circ})}} A=7sin⁡(48.9∘)sin⁡(43.2∘)A = \frac{7 \sin(48.9^{\circ})}{\sin(43.2^{\circ})} A=7.7A = 7.7

Great! We solved for a,Aa, A and CC! Let's check our ratios, for they should all equal each other!

A: 7.7sin⁡(48.9∘)=10.22\frac{7.7}{\sin(48.9^{\circ})} = 10.22

B: 7sin⁡(43.2∘)=10.22\frac{7}{\sin(43.2^{\circ})} = 10.22

C: 10.2sin⁡(87.9∘)=10.21\frac{10.2}{\sin(87.9^{\circ})} = 10.21

Note: CC is off by 0.010.01, which is not enough to matter for this problem!

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