**THE LAWS OF SINES AND COSINES**

Example 1:

For the oblique triangle with measures

a = 12,ftb = 31, andftA = 20.5, find the remaining sides and angles. Round all calculations to two decimal places.^{o}Assume angles

,A, andBare opposite sides labeledC,a, andbrespectively.cHINT: Make a sketch!!!

Given two sides and one angle opposite one of the sides, we can use the

Law of Sines.

Solve for Angle B:Specifically, we'll write

12 sin B = 31 sin 20.5^{o}Please note that

is positive. We know that the value of sine is positive for QI and QII angles!sin BNOTE: Since the sum of interior angles in a triangle is

and QI and QII angles are less than180^{o}, we could get two solutions for angle180^{o}, one acute and one obtuse angle.BUsing the calculator, we find

Then, the QI angle is

B_{1}.64.78^{o}Since

B_{1}_{ }is also the reference angle, we find the QII angle to be

B_{2}=180^{o}64.78^{o}115.22^{o}Both of these angles could be solutions for the given triangle! As long as angle

is not negative, we will have a triangle with two solution sets.C

Solve for Angle C:We know that

A = 20.5,^{o}B_{1}, and64.78^{o}B_{2}.115.22^{o}Then

and

Since both measures for angle

Care positive the triangle indeed has TWO solution sets.

Solve for Side c:Given , let's find side using the

Law of Sinessince we have two sides and one opposite angle:

12 sin 94.72^{o}= c_{1 }sin 20.5^{o}Given , let's find side by using the

DO NOT CALCULATE the individual values of the trigonometric ratios, instead use the EXACT value as shown below.Law of Sines.

12 sin 44.28^{o}= c_{2}sin 20.5^{o}

DO NOT CALCULATE the individual values of the trigonometric ratios, instead use the EXACT value as shown below.

The triangle has the following solutions:Given

A = 20.5,^{o}, anda = 12 ft,b = 31 ftwe found

B_{1 },64.78^{o}, andC_{1}94.72^{o}34.15 ftc_{1}or

B_{2 },115.22^{o}, andC_{2}44.28^{o}23.92 ftc_{2}

Example 2:

For the oblique triangle with measures

a = 22,inb = 12, andin, find the remaining sides and angles. Round all calculations to two decimal places.A = 42^{o}Assume angles

,A, andBare opposite sides labeledC,a, andbrespectively.cHINT: Make a sketch!!!

Given two sides and one angle opposite one of the sides, we can use the

Law of Sines.

Solve for Angle B:Specifically, we'll write

22 sin B = 12 sin 42^{o}Please note that

is positive. We know that the value of sine is positive for QI and QII angles!sin BNOTE: Since the sum of interior angles in a triangle is

and QI and QII angles are less than180^{o}, we could get two solutions for angle180^{o}, one acute and one obtuse angle.BUsing the calculator, we find

Then, the QI angle is

B_{1}.21.41^{o}Since

B_{1}_{ }is also the reference angle, we find the second-quadrant angle to be

B_{2}=180^{o}21.41^{o}158.59^{o}Both of these angles could be solutions for the given triangle! As long as angle

is not negative, we will have a triangle with two solution sets.C

Solve for Angle C:We know that

A = 42,^{o}B_{1}, and21.41^{o}B_{2}.158.59^{o}This is

NOTa solution, since a triangle hasNOnegative interior angles. Thus, is alsoNOTa solution.We have just proven that this particular triangle only has ONE solution set!

Solve for Side c:Given

C 116.59, let's find side^{o}using thecLaw of Sinessince we have two sides and one opposite angle:

DO NOT CALCULATE the individual values of the trigonometric ratios, instead use the EXACT value as shown below.

c 29.40 in

The triangle has the following solutions:Given

A = 42,^{o}, anda = 22 inb = 21 in,we found

B 21.41,^{o}, andC 116.59^{o}c 29.40in

Example 3:

For the oblique triangle with measures

a = 15,inchesb = 25andinches,A = 85, find the remaining sides and angles. Round all calculations to two decimal places.^{o}Assume angles

,A, andBare opposite sides labeledC,a, andbrespectively.cHINT: Make a sketch!!!

Given two sides and one angle opposite one of the sides, we can use the

Law of Sines.

Solve for AngleB:Specifically, we'll write

15 sin B = 25 sin 85^{o}Please note that

is positive. We know that the value of sine is positive for QI and QII angles!sin BNOTE: Since the sum of interior angles in a triangle is

and QI and QII angles are less than180^{o}, we could get two solutions for angle180^{o}, one acute and one obtuse angle.BUsing the calculator, we find

But the calculator tells us "Domain Error".

Investigating, we find

The result shows that

is greater than 1. However, the value ofsin Boscillates only betweensin Band1.1

Therefore, we must conclude that the triangle with the given measurements has NO solutions.Below is a more accurate picture of the given information:

Example 4:

Solve the following oblique triangle. Round all calculations to two decimal places.

Let's use the

Law of Cosinesbecause we do not have at least one side and its opposite angle which is necessary to use theLaw of Sines.

Solve for Angle A:Use .

Note that

was not changed to a decimal approximation in the interest of more exact solutions.493/532There cannot be another solution because the only other quadrant in which

is positive is QIV, but there the angles are greater thancosine.180^{o}

Solve for Angle B:We'll use the

Law of Cosinesagain. Specifically, we will use.Note that

was not turned into a decimal approximation!101/224

Solve for Angle C:We know that

A 22.5and^{o}B116.80^{o}Let's use the fact that the sum of interior angles in a triangle is

.180^{o}

C 180^{o}22.08^{o}116.8^{o}= 41.12^{o}

The triangle has the following solutions:Given

,a = 8 ft, andb = 19 ft,c = 14 ftwe found

,A 22.08^{o}, andB 116.8^{o}C41.12^{o}

Example 5:

Solve the oblique triangle pictured below. Round all calculations to two decimal places.

Given two sides and one angle opposite one of the sides, we could use the

Law of Sines.However, let's instead use theLaw of Cosinesto show that this law can be used as well.

Solve for Side b:Let's use .

Note that the radicand was not changed to a decimal approximation before taking its square root! It is better to keep the numbers as exact as possible for as long as possible.Disregarding the negative value above (distances are never negative!!!), we get

b1.03 ft

Solve for Angle A:Let's use using the UNROUNDED value for Side

.b

Please note that the fraction above was not turned into a decimal approximation. It was typed into the calculator using appropriate grouping symbols to keep the solution as exact as possible.

A160.27^{o}There cannot be another angle because the only other quadrant in which

is negative is Q III, where the angles are greater thancos A180.^{o}

Solve for Angle B:We know that

A 160.27and^{o}C =.10^{o}Let's use the fact that the sum of interior angles in a triangle is

180^{o}.^{}

B180^{o}160.27^{o}10^{o}= 9.73^{o}

The triangle has the following solutions:Given

,a = 2 ft, andc = 1 ft10,^{o}we found

A,160.27^{o}, andB 9.73^{o}b 1.03 ft