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Teachers & SchoolsThe law of sines is all about proportions, so make sure to keep yours in order.

Functions | Prove and apply trigonometric identities |

Language | English Language |

Right Triangles and Trigonometry | Law of Sines and Law of Cosines |

Dr. Dorito tells security that two of his lengths are 3.2 and 4, and that his angle

A is 60 degrees. Given Angle-Side-Side, to figure out the rest

of the angles and side lengths, they'll use the law of sines.

Sine A over a equals sine B over b.

We can plug in A as 60 degrees, and side length A as 4.

The only other side we have is 3.2, so that'll be b.

We have sine of 60 degrees over 4 equals sine of the angle at point B over 3.2, the side

length of b. To solve for angle B, we can multiply both

sides by 3.2. We want B, not sine B. So if we take the inverse sine of each side, we

see that B is around 43.9 degrees. Wonderful. Now we have two out of three angles

and two out of three sides. Knowing that all the angles in a triangle

add up to 180° allows us to find the last angle.

Substitute in the angles we know. A as 60 degrees and B as 43.9 degrees.

To solve for C, we can subtract 60 and 43.9 from both sides, to get C equals 180 degrees

minus 60 degrees plus 43.9 degrees... ...to get that C equals 76.1 degrees.

One side to go. We can use either the ratio between sine 60 degrees and 4 or sine 43.9

degrees and 3.2. Let's use the ratio between sine 60 degrees

and 4. Substitute what we know. We're looking for

the last side, c. To solve for side length c, we can rearrange

the equation by cross-multiplying. Sine sixty degrees times c equals 4 times

sine 76.1 degrees. To isolate c, we can divide both sides by sine sixty degrees...to get

that c equals around 4.5. Double-checkin' time. Do we know all of Dr.

Dorito's angles and all the sides? Yup, sure do. All three side lengths and all

three angles. Now they just have to make sure he's not smuggling

any zesty taco flavoring across the border...