# Sines

#### The relation between sines and chords

In this section we’ll only consider sines of angles between 0° and 90°. In the section on trigonometric functions, we’ll define sines for arbitrary angles.

A sine is half of a chord. More accurately, the sine of an angle is half the chord of twice the angle.

The bow and arrow diagram

Consider the angle BAD in this figure, and assume that AB is of unit length. Let the point C be the foot of the perpendicular dropped from B to the line AD. Then the sine of angle BAD is defined to be the length of the line BC, and it is written sin BAD. You can double the angle BAD to get the angle BAE, and the chord of angle BAE is BE. Thus, the sine BC of angle BAD is half the chord BE of angle BAE, while the angle BAE is twice the angle BAD. Therefore, as stated before, the sine of an angle is half the chord of twice the angle.

The point of this is just to show that sines are all that difficult to understand. (Whoops, that’s a slip! I meant to write “not all that difficult to understand.”)

#### The meaning of the word “sine”

The Sanskrit word for chord-half was jya-ardha, which was sometimes shortened to jiva. This was brought into Arabic as jiba, and written in Arabic simply with two consonants jb, vowels not being written. Later, Latin translators selected the word sinus to translate jb thinking that the word was an arabic word jaib, which meant breast, and sinus had breast and bay as two of its meanings. In English, sinus was imported as “sine.”

This word history for sine is interesting because it follows the path of trigonometry from India, through the Arabic language from Baghdad through Spain, into western Europe in the Latin language, and then to modern languages such as English and the rest of the world.

#### Sines and right triangles

We can use properties of similar triangles to relate sines to right triangles. In the figure above the triangle ABC is a right triangle with a right angle at angle C and a hypotenuse of length 1. Consider a similar right triangle AB'C' with a hypotenuse of arbitrary length.(If your web browser is Java-enabled, you can drag the points B' to change the size of the right triangle AB'C'.)

Since the triangles are similar, the ratio BC to AB equals the ratio B'C' to AB'. But AB equals 1. Hence,

BC = B'C' / AB'

but BC = sin A, so

sin A = B'C' / AB'

This result is most easily remembered as the sine of an angle in a right triangle equals the opposite side divided by the hypotenuse: #### The standard notation for a right triangle

For a while we’ll be looking mainly at right triangles, so it would be useful to use a standard notation for the angles and sides of these triangles.

Consider a right triangle ABC with a right angle at C. We’ll generally use the letter a to denote the side opposite angle A, the letter b to denote the side opposite angle B, and the letter c to denote the side opposite angle C, that is, the hypotenuse.

With this notation, sin A = a/c, and sin B = b/c.

Next we’ll look at cosines. Cosines are just sines of the complementary angle. Thus, the name “cosine” (“co” being the first two letters of “complement”). For triangle ABC, cos A is just sin B. #### Exercises

28. In a right triangle the hypotenuse c = 15 inches, and the sine of one angle A, is sin A = 2/5. Find a, the side opposite A, and find b, the remaining side.

44. In a right triangle B = 55° 30', and b = 6.05. Find c and a.

191. If the height of a gable end of a roof is 22.5 feet and the rafters are 30 feet 8 inches long, at what angle do the rafters slope, and how wide is the gable end at the base?

194. The top of a ladder 50 feet long rests against a building 43 feet from the ground. At what angle does the ladder slope, and what is the distance of its foot from the wall?

#### Hints

28. The hypotenuse c is 15". Since sin A = a/c, therefore a = c sin A. That gives you a. Next use the Pythagorean theorem to find b knowing a and c.

44. Since sin B = b/c, you can determine c. Once you’ve got b and c, you can determine a by the Pythagorean theorem.

191. A gable end ABD of a roof is an isosceles triangle with the base being the width of the house, and the two equal sloping sides the rafters at the end of the roof. If you drop a perpendicular from the apex B of the triangle, you’ll get two congruent right triangles, ABC and DBC. Since you know two sides of the right triangle ABC, you can compute the third by using the Pythagorean theorem. You can use sines to determine the angle of slope, since sin A = BC/AB = 22.5'/30'8" = 0.7337. To find the angle A, you’ll need what’s called the arcsine of 0.7337.

The arcsine function is inverse to the sine function, and your calculator can compute them. Usually there’s a button on the calculator labeled “inv” or “arc” that you press before pressing the sin button. Then you’ll have the angle. Your calculator can probably be set to either degree mode or radian mode. If it’s set to degree mode, then you’ll get the angle in degrees; and if it’s set to radian mode, then you’ll get the angle in radians. Always be sure you know which mode your calculator’s set to.

194. Draw a triangle ABC as above. You know the hypotenuse c and the vertical side a. The distance b can be found by the Pythagorean theorem. Just take the square root of c2 – a2. You can find the slope, that is, angle A, using sines. You know sin A = a/c = 43/50 = 0.86. As in problem 191, use arcsin to find the angle A.