Trigonometry: Right triangles and radian measure
The study of similarity within right triangles provides the background for the study of trigonometry. The concept is extended when the triangles are put on the coordinate plane, and the triangles are located somewhere other than the first quadrant. A new measure for measuring angles, radian measure, is based strictly on the measurements of the geometric figures involved.
Right triangle trigonometry
Trigonometry started as an application of similar right triangles. The three basic functions—sine, cosine, and tangent—are remembered with the mnemonic SOHCAHTOA: The sine ratio (S) is the ratio of the lengths of the side opposite (O) to the acute angle to the hypotenuse (H); the cosine (C) is the ratio of the adjacent (A) side to the hypotenuse (H); and the tangent (T) is the ratio of the opposite (O) side to the adjacent (A) side. Each ratio is abbreviated with three letters: The sine ratio is sin, the cosine ratio is cos, and the tangent ratio is tan.
In the diagram shown, angle C is a right angle. The sine of angle A, written sin(A), is 
the cosine of angle A, cos(A), is 
and the tangent of angle A, tan(A), is 
Similarly, sin(B) = 
cos(B) = 
and tan(B) = 
Angles A and B are complementary angles. In fact, the word cosine comes from complement of the sine. For any acute angle, it will always be the case that sin(A) = cos(90 − A).
For questions 1–3, find the length of the missing side in each right triangle.
1. 
3. 
2. 
Find the measure of angle θ, to the nearest degree, for questions 4–6.
4. 
6. 
5. 
Meghan was hiking when she came to a flat region and saw a hill in the distance. Being a bit of a math lover, she happens to have a inclinometer in her backpack. She measures the angle of elevation to the top of the hill to be 18.3°. She walks directly toward the hill (in measured strides) another 400 ft. She measures the angle of elevation to the top of the hill to be 43.7°. She knows that she can now calculate the height of the hill when she gets back home.
7. Determine, to the nearest foot, the height of the hill.
Special right triangles
There are two triangles that have special importance in the study of trigonometry: the 30–60–90 and 45–45–90 triangles. The 30–60–90 can be created by dividing an equilateral triangle in two, while the 45–45–90 (isosceles right triangle) comes from cutting a square in two.
If each side of the equilateral triangle has length 2a, then the length of the segment from A to midpoint C is a. Use the Pythagorean theorem to show that BC = 
If each side of the square has length a, the length of the diagonal of the square is 
The values for the three trigonometric functions for the acute angles of these triangles are shown in the following table.
The unit circle: First quadrant
The unit circle is a circle with its center at the origin and with radius of length 1. When the right triangle is placed inside this circle, as shown in the figure, important relationships are determined.
With the acute angle at the origin measuring θ, OA = cos(θ), and AB = sin(θ). The coordinates of point B are [cos(θ), sin(θ)]. The equation of the unit circle is x2 + y2 = 1, and the coordinates for point B translate to the most important trigonometric identity [cos(θ)]2 + [sin(θ)]2 = 1. This is more commonly written as cos2(θ) + sin2 (θ) = 1 and is one of the three Pythagorean identities.
∆OAB ~ ∆OCT because they share angle θ and they each have a right angle. Recalling that corresponding sides of similar triangles are in proportion, three very important relationships can be found.
First, 
OA = cos(θ), and OC = 1. The proportion becomes 
= CT. However, CT is a segment on the line drawn tangent to the circle at point C. It is for this reason that CT = tan(θ). Therefore, 
The second important proportion from these similar triangles is 
and OA = cos(θ). OT is a segment on the line which passes through the circle twice (extend the ray through O to get the second point of intersection). Such a line is called the secant line. Therefore, OT, which is the reciprocal of cos(θ), is called the secant of θ, abbreviated sec(θ). Therefore, 
∆OCT is a right triangle, so the Pythagorean theorem applies to it. OC2 + CT2 = OT2. Substituting the new trigonometric values just learned, this equation becomes the second Pythagorean identity, 1 + tab2(θ) = sec2(θ).
because they are both perpendicular to the y-axis. Therefore, the measure of angle DSO must also be θ, as it is an alternate interior angle to angle BOA, and ∆OCT ~ ∆SDO. The ratio of corresponding sides yields two new trigonometric functions. 
With OS = 1 and SD being the line drawn tangent from the angle complementary to angle θ, 
is called the cotangent function, cot(θ), so that 
By the same token, OS is called the cosecant of θ, csc(θ) so that 
Finally, ∆SDO is a right triangle and OD2 + DS2 = OS2, or 1 + cot2(θ) = csc2(θ), which is the third Pythagorean identity.
Complete the chart.
Given the acute angle Z with 
find the value of the trigonometric functions specified in questions 10–14.
10. sec(Z)
11. sin(Z)
12. csc(Z)
13. tan(Z)
14. cot(Z)
Given the acute angle X with 
find the value of the trigonometric functions specified in questions 15–19.
15. sin(X)
16. cos(X)
17. sec(X)
18. tan(X)
19. cot(X)
The unit circle—beyond the first quadrant
Angles of rotation become an issue when angles move beyond the first quadrant. When drawn in standard position, the initial side of the angle is the positive x-axis. If the terminal side is drawn in a counterclockwise manner from the initial side, the angle is said to have positive measure; if drawn in a clockwise manner, the angle has negative measure. Angles whose terminal sides are the same ray are called coterminal angles. For example, an angle with measure 130° and an angle with measure −230° are coterminal. These same angles are coterminal with angles having measures 490°, 850°, 1210°, −590°, and −950°. It is possible to have angles with measure greater than 360° (think about a car spinning on ice—“doing a 360”—more than one revolution yields an angle that is more than 360°).
When the terminal side of θ goes beyond the first quadrant, the rules for opposite, adjacent, and hypotenuse need to be reconsidered. For example, when θ = 90°, the coordinates for point B are (0, 1). Therefore, cos(90) = 0 and sin(90) = 1. csc(90) is also 1, while sec(90) and tan(90) are both undefined. (Stop a moment to think about this from an algebraic perspective, 
is undefined. From a geometric perspective, 
so there are no points of intersection.) Angles that terminate on one of the axes are called quadrantal angles.
If θ terminates within one of the quadrants, reflexive symmetry is used from a corresponding point in the first quadrant to determine the trigonometric values.
In each case, when point B is reflected back into the first quadrant, a triangle congruent to ∆OAB is formed. The acute angle in the first quadrant is called the reference angle for θ. It is imperative that you notice that the reference angle is always the acute angle that the terminal side of θ makes with the x-axis.
Name two coterminal angles (one with positive measure and one with negative measure) for each of the angles in questions 1–3.
1. 119°
2. 437°
3. −97°
Find the measures of the reference angles for questions 4–7.
4. 95°
5. 190°
6. 290°
7. 517°
For questions 8–10, express each trigonometric statement as a function of a positive acute angle.
8. tan(217°)
9. sin(319°)
10. cos(129°)
For questions 11–13, determine the coordinates for each of the points on the unit circle.
11. 120°
12. 225°
13. 330°
For questions 14 and 15, find the specified trigonometric value.
14. Given 
with ∠A terminating in quadrant III, find cos(A).
15. Given 
with ∠B terminating in quadrant II, find tan(B).
Radian measure
By this point in your education, you probably have had many experiences dealing with different units of measurement. There is standard (inches, feet, yards, miles) versus metric (centimeter, meter, kilometer), as well as Fahrenheit versus Celsius. The use of the degree as a measure of angles goes back to the Babylonians and a calendar with 360 days (a five-day religious celebration at the end of the year kept their calendar relatively accurate for their short time as a power in history).
A more accurate measure of angle measure (despite its lack of usage among the general public) is the radian. By definition, the radian measure of a <central> angle is the ratio of the arc formed by the angle to the length of the radius. A key piece of this definition is that the radian has no units. Arc length and radius will be measured in the same units (whether standard or metric) so will cancel each other within the ratio.
The conversion between degrees and radians is best considered when using a circle with radius 1 (although this is not necessary, as a dilation will create proportionally large radii and arcs without changing angles). A complete revolution of the circle is 360°, but it is also an arc of length 2π radians. Half a revolution is 180° or π radians. It is usually this reduced set of numbers that is used to convert from one measurement to the other: 
If θ is the radian measure of the central angle, r is the length of the radius, and s is the length of the arc, the formula 
becomes s = rθ.
Convert each of the angle measures in questions 1–3 to radian mode.
1. 72°
2. 140°
3. 315°
Convert each of the angle measures in questions 4–6 to degree mode.
4. 
5. 
6. 
Given the information in questions 7 and 8, solve for the specified values.
7. In a circle with radius 8 cm, a central angle forms an arc with length 20 cm. Find the radian measure of the central angle.
8. A central angle with measure 
forms an arc with length 4 cm. Find the length of the radius of the circle.
Basic trigonometric identities
Recapping the chapter so far, there are 21 relationships that have been established with regard to the trigonometric functions.
Reciprocal identities
Pythagorean identities
Reflections
This can be extended for the other functions as well as other quadrants.
In addition to these relationships, there is the issue of counterclockwise versus clockwise rotations of the terminal side of the angle.
Rotational
Look at the diagrams for angles terminating in the different quadrants to see this.
There are a number of other important formulas that you will need to be comfortable with applying.
Addition and subtraction formulas
Double angles
Given 
and 
find the answers for questions 1–3.
1. sin(A − B)
2. cos(A − B)
3. tan(A − B)
Given 
find the answers for questions 4–6.
4. tan(45 + G)
5. sin(G − 60)
6. sin(60 − G)
Given 
find the answers for questions 7–9.
7. 
8. 
9. 
Area of a triangle
The area of ∆ABC is 
In right ∆ADC, 
so that h = AC sin(A). Substitute this into the formula for the area of the triangle and get the result that the area of ∆ABC is 
. In general, the area of a triangle is equal to one-half the product of the lengths of two sides of a triangle and the sine of the included angle.
Using the information given, solve the following.
1. Given ∆ABC, AB = 20 in, AC = 30 in, and m 
A = 47°, find the area of ∆ABC to the nearest square inch.
2. Given ∆WHY with 
and m H = 123°, find the area of ∆WHY to the nearest square foot.
3. The area of ∆KLM is 8693 cm2. If 
and 
find the measure of angle K to the nearest degree.
4. The area of obtuse ∆GHJ is 71,365 cm2. If 
and 
find the measure of the obtuse angle G to the nearest degree.
Law of sines
The area of ∆ABC can be computed with the product: 
It can also be computed with the product 
or 
Since all three expressions represent the area of the same triangle, they must be equal.
Divide all three expressions by 
to get the equation that is called the law of sines: 
Using the convention of naming the side of a triangle with the lowercase letter matching the vertex of the angle, this becomes the more familiar form of the equation: 
It is important to note that the proportion created by any two fractions in the law of sines involves two sides and two angles, and that the orientation of these sides and angles fits the patterns AAS, ASA, or SSA, which you may remember from your study of geometry.
Solve the following using the information given.
1. Given ∆ACF with AC = 20 cm, AF = 40 cm, and m C = 73.2°, find m F to the nearest tenth of a degree.
2. Given ∆KLM with KL = 90.2 ft, LM = 131.4 ft, and m K = 59°, find m M to the nearest tenth of a degree.
3. Given ∆XYZ with m X = 67°, m Y = 47°, and XY = 50.2 mm, find XZ and YZ to the nearest tenth of a centimeter.
4. Given ∆QRS with m Q = 37.2°, m R = 43.1°, and QR = 11.7 mm, find QS and RS to the nearest tenth of a centimeter.
5. Given ∆YES with m Y = 47.1°, m E = 73.6°, and YS = 61.7 mm, find ES and EY to the nearest tenth of a centimeter.
6. Given ∆QED with m Q = 71.2°, m E = 63.1°, and QD = 53.2 mm, find QE and ED to the nearest tenth of a centimeter.
Ambiguous case
Given C and BC = 30 cm with m C = 40°, what is the length of the shortest segment that can be drawn from point B to create a triangle?
If arcs are drawn from vertex B, as shown in the accompanying diagram, there is one arc that intersects the ray of angle C only once.
The point of intersection for this arc is the foot of the perpendicular from B to the ray.
Hopefully, this makes sense to you because the shortest distance from a point to a line is along the perpendicular. This is why you are told to stand up “straight” when someone is measuring your height.
If point A is the point at which the perpendicular intersects the side of the ray, then ∆ABC is a right triangle and BA = 30 sin(40).
If an arc is drawn with a length AB that is less than the length of the perpendicular, BC sin(C), the arc will never intersect the side of the ray. It would not be possible to create a triangle with the given measurements.
If an arc is drawn with a length AB that is greater than BC, the arc will intersect the ray somewhere to the right and one triangle can be constructed.
If an arc is drawn with a length AB that is greater than BC sin(C) but less than BC [i.e., BC sin(C) < AB < BC], then there will be two triangles that can be constructed. It is this scenario that is referred to as the ambiguous case.
The situations described above apply when the angle in question is an acute angle. If the given angle is either a right angle or an obtuse angle, the length of the arc drawn from the endpoint of the segment must be longer than the segment itself or the arc will not intersect the other side of angle C.
Using the information provided, determine how many triangles, if any, can be constructed for each of the following.
1. In ∆KAT, KA = 68 in, KT = 34 in, and m A = 30°.
2. In ∆DAY, DA = 126 yd, YD = 63 
yd, and m A = 60°.
3. In ∆TAD, TA = 82 mm, TD = 44 mm, and m D = 40°.
4. In ∆NET, TE = 32 cm, TN = 44 cm, and m E = 130°.
5. In ∆ARG, RA = 95 m, RG = 67 m, and m A = 110°.
6. In ∆MBA, MA = 72 cm, AB = 60 cm, and m M = 50°.
7. In ∆EDU, EU = 125 mm, DU = 114 mm, and m E = 57°.
Law of cosines
The law of sines is used when the information available for the triangle fits one of the patterns ASA, AAS, or SSA. When the information available for the triangle fits the pattern SSS or SAS, the law of cosines is used to find the missing information about the triangle. The formula for the law of cosines is
p2 = q2 + r2 − 2 qrcos(P)
The key to this formula is to realize that when the information fits the SAS pattern, it is the third side of the triangle that is determined first. When the information is of the SSS pattern, it is an angle that is determined first. In the formula for the law of cosines, the side and the angle opposite that side are at the beginning and end of the formula.
The temptation is to subtract 503.04 from 540.25, but this is not mathematically correct. In the same way that one cannot simplify 5 − 3x because the terms are not “like” terms, 540.25 and 503.04 cos(Y) are not like terms.
The law of sines or the law of cosines can be used to determine m S and m P if you are directed to do so, although the law of sines would be less work computationally.
The law of sines and the law of cosines are used in the resolution of vectors. Vectors are represented as directed line segments (they have length and direction) and the sum of two vectors is accomplished using the parallelogram method. That is, the vectors are placed together at their “tails” (endpoints) with the direction of the vector indicated with an arrow as shown in the accompanying diagram.
A parallelogram is constructed from these two sides, and the diagonal of the parallelogram drawn from the original set of tails is called the resultant, or net impact of the two vectors. The resulting picture is the diagram for a vector sum.
Solve the following.
1. Given ∆ NEC with NC = 20 cm, NF = 40 cm, and m N = 73.2°, find EC to the nearest centimeter.
2. Given ∆ MVP with MV = 90.2 ft, MP = 131.4 ft, and m M = 59°, find VP to the nearest foot.
3. Given ∆XYZ with XY = 67 mm, YZ = 47 mm, and XY = 50 mm, find m X to the nearest tenth of a degree.
4. Given ∆PTS with PT = 37.2 mm, PS = 43.1 mm, and TS = 11.7 mm, find m T to the nearest tenth of a degree.
5. Vectors with magnitudes 83 N and 75 N act on an object at an angle of 40° to each other. Find the magnitude of the resultant force to the nearest tenth of a newton.
6. Vectors with magnitudes 100 N and 120 N act on an object with a resultant force of 160 N. Find, to the nearest degree, the angle between the two original forces.
7. Vectors with magnitudes 50 N and 70 N act on an object with a resultant force of 85 N. Find, to the nearest tenth of a degree, the angle between the resultant and the larger force.