# mem 104 2022

```MEM 104
ALGEBRA AND PLANE
TRIGONOMETRY
LINDY LOU C. PASCO
Presenter
TOPIC
Solutions of Right Triangles
Solutions of Right Triangles
The process of finding the missing measurements is known as solving the triangle
If the triangle is a right triangle, then one of the angles is 90&deg;. Therefore, you can solve the right triangle if you are given the
measures of two of the three sides or if you are given the measure of one side and one of the other two angles .
If the triangle is a right triangle, you can use simple trigonometric ratios to find the missing parts.
A. Solving a right triangle given the measure of the two parts; the length of the hypotenuse and the length of one leg
Example:
Triangle BCA is right-angled at C. If π = 23 and π = 17 , find ∠ π΄, ∠π΅ and π . Express your answer into two decimal places.
Solution:
a. Side b is the adjacent side of ∠π΄; c is the hypotenuse of right triangle BCA. Use CAH, that is
ππππππππ‘
cos π =
βπ¦πππ‘πππ’π π
Sketch the figure
π
cos π΄ =
π
17
cos π΄ =
23
cos π΄ = 0.7391
π΄ ≈ 42.3400
We use scientific calculator to find an angle whose cosine value is 0.7391
b. Since in part (a), it was already found that ∠π΄ ≈ 42.34π then,
∠π΅ = 90π − 42.34π
∠π΅ ≈ 47.66π
c. Find a by using the Pythagorean Theorem
π2 + π 2 = π 2
π2 + (17)2 = (23)2
π2 = 529 − 289
π2 = 240
π =
240
π ≈ 15.49
The complete set of solutions of the given
triangle are:
∠π΅ ≈ 47.66π
∠π΄ ≈ 42.34π
∠πΆ = 900
π ≈ 15.49
π = 17
π = 23
B. Solving a right triangle given the length of the hypotenuse and the measure of one acute angle
Example:
Triangle BCA is right-angled at C, if π = 27 and ∠ π΄ = 58π , find ∠ π΅, b, and a.
Solution:
a. To find B, since B and A are complementary angles, then
∠π΅ + ∠π΄ = 90π
∠π΅ = 90π − 58π
∠π΅ = 32π
b. To find b, since b is the adjacent side of ∠π΄ and c is the hypotenuse of right βπ΅πΆπ΄ , then use CAH
ππππππππ‘
cos π =
π = 27 cos 58π
βπ¦πππ‘πππ’π π
π
cos π΄ =
π
cos 58π =
π
27
π = 27(0.5299)
π ≈ 14.31
c. To find a, since a is the opposite side of ∠π΄ and c is the hypotenuse of right βπ΅πΆπ΄ , then use SOH
πππππ ππ‘π
sin π =
βπ¦πππ‘πππ’π π
π
sin π΄ =
π
π
sin 58 =
27
π
π = 27 sin 58π
The complete set of solutions of the given
triangle are:
∠π΅ = 320
∠π΄ = 580
∠πΆ = 900
π = 27(0.8480)
π ≈ 22.9
π ≈ 22.9
π ≈ 14.31
π = 27
C. Solving a right triangle given the length of one leg and the measure of one acute angle
Example:
Solution:
Triangle ACB is right-angled at C. If ∠ π΄ = 63π and a= 11 cm, find ∠ π΅, b, and c.
a. To find ∠B, take note that ∠ B and ∠ A are complementary angles. Then,
∠ B + ∠ A= 90π
∠ B = 90π − 63π
∠ B = 27π
b. To find b, since b is the adjacent side and a is the opposite side of ∠π΄, then use TOA
πππππ ππ‘π
11
π
tan π =
tan 63 =
ππππππππ‘
π
π
tan π΄ =
π tan 63π = 11
π
tan 63π
11
=
π
π(1.9626) = 11
11
π=
1.9625
π ≈ 5.60 ππ
c. To find c, since c is the hypotenuse and a is the opposite side of ∠π΄, then use SOH
πππππ ππ‘π
sin π =
βπ¦πππ‘πππ’π π
The complete set of solutions of the given
triangle are:
∠π΅ = 270
∠π΄ = 630
∠πΆ = 900
π
sin π΄ =
π
sin 63π =
11
π
π = 11
π ≈ 5.60 ππ
π ≈ 12.35 ππ
π sin 63π = 11
π(0.8910) = 11
11
π=
0.8910
π ≈ 12.35 ππ
D. Solving a right triangle given the length of the two legs
Example:
Triangle ACB is right-angled at C. If a =18.5 cm and b= 14.2 cm, find c, ∠ π΄ πππ ∠ π΅.
Solution:
a. To find c, use Pythagorean Theorem
π 2 = π2 + π2
π 2 = (18.5)2 +(14.2)2
b. To find ∠π΄ , since a and b are opposite and adjacent side of ∠π΄ respectively, then use TOA
π 2 = 342.25 + 201.64
π2
= 543.89
π = 543.89
π ≈23.32 cm
πππππ ππ‘π
tan π =
ππππππππ‘
π
tan π΄ =
π
tan π΄ =
18.5
14.5
tan π΄ = 1.3028
π΄ ≈ 52π
We can use our scientific calculator to
find an angle whose tangent is 1.3028
c. Based on the fact that ∠ A and ∠ B are complementary, the measure of angle
∠ B is 90π − 52π = 38π
The complete set of solutions of the given
triangle are:
∠π΅ = 380
∠π΄ ≈ 52π
∠πΆ = 900
π = 18.5 ππ
π = 14.2 ππ
π ≈23.32 cm
Law of Sines
Learning Objectives
1. Use the Law of Sines to solve oblique triangles.
2. Find the area of an oblique triangle using the sine function.
3. Solve applied problems using the Law of Sines.
Using the Law of Sines to Solve Oblique Triangles
Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so,
we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations:
1. ASA (angle-side-angle) We know the measurements of two angles and the included side.
2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles.
3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides.
Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles.
Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two
ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 10.6.
β
π
β
π
Using the right triangle relationships, we know that sin α = and sin β = . Solving both equations for β gives two different expressions for β.
β= π sin πΌ and β = π sin π½
We then set the expressions equal to each other.
Similarly, we can compare the other ratios.
Collectively, these relationships are called the Law of Sines
Law of Sines
Given a triangle with angles and opposite sides labeled as in Figure 10.7, the ratio of the measurement of an angle to the length of its opposite side will be
equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented
symbolically two ways.
To solve an oblique triangle, use any pair of applicable ratios.
Solving for Two Unknown Sides and Angle of an AAS Triangle
Solve the triangle shown in Figure 10.8
Solution
The three angles must add up to 180 degrees. From this, we can determine that
π½ = 180&deg; − 50&deg; − 30&deg;
= 100&deg;
To find an unknown side, we need to know the corresponding angle and a known ratio. We know that angle πΌ = 50&deg; and its corresponding side π= 10. We can use the following proportion from the
Law of Sines to find the length of π.
π ππ 500
sin 300
=
sin πΌ sin πΎ
10
π
=
π
π
π sin 50π = 10 sin 300
π =
10 sin 300
π ππ 500
π ≈ 6.5
Similarly, to solve for π, we set up another proportion.
sin πΌ sin π½
=
π
π
sin 500
π ππ 1000
=
10
π
π sin 500 = 10 sin 1000
10 sin 1000
π=
sin 500
π ≈ ππ. π
Therefore, the complete set of angles and sides is
Example 10.2
Solve the given triangle using the Law of Sines. Round lengths to the nearest tenth and angle measurements to the nearest degree.
Find side b
Since we are given angle A and side a we will use these in our ratio setup to find side b.
π
π
=
sin π΄ sin π΅
12
π
=
sin 70π sin 55π
A = 70&deg;, B = 55&deg;, and a = 12
Solution:
Find angle C
The sum of the angles of a triangle must equal 180&deg; so C would be the difference
between 180&deg; and the sum of the other two angles.
A + B + C = 180&deg;
70&deg; + 55&deg; + C = 180&deg;
C = 180&deg; – 125&deg;
C = 55&deg;
12 sin 55π
=π
sin 70π
π ≈ 10.5
Find side c
Even though we just found side b you will still want to use the measurement that were given to us in the problem. The
length of side b is an approximated value and will not produce as good of an answer as the known values for side a and
angle A.
π
π
=
sin π΄ sin πΆ
12
π
=
sin 70π sin 55π
12 sin 55π
=π
sin 70π
Since angles B and C are the same the lengths of sides b and c will also be the same.
π ≈ 10.5
Using The Law of Sines to Solve SSA Triangles
We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as
an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two
solutions, or even no solution.
Possible Outcomes for SSA Triangles
Oblique triangles in the category SSA may have four different outcomes. Figure 10.10 illustrates the solutions with
the known sides a and b and known angle α.
Solving an Oblique SSA Triangle
Solve the triangle in Figure 10.11 for the missing side and find the missing angle measures to the nearest tenth.
Solution
Use the Law of Sines to find angle β and angle γ, and then side c. Solving for β, we
have the proportion
However, in the diagram, angle β appears to be an obtuse angle and may be greater than 90&deg;. How did we get an acute angle, and how do we find the measurement of β ? Let’s
investigate further. Dropping a perpendicular from γ and viewing the triangle from a right angle perspective, we have Figure 10.12. It appears that there may be a second triangle that
will fit the given criteria.
The angle supplementary to β is approximately equal to 49.9&deg;, which means that β = 180&deg; −
49.9&deg; = 130.1&deg;. (Remember that the sine function is positive in both the first and second
quadrants.) Solving for γ, we have
γ = 180&deg; − 35&deg; − 130.1&deg; ≈ 14.9&deg;
We can then use these measurements to solve the other triangle. Since γ′ is supplementary to γ, we
have
γ′ = 180&deg; − 35&deg; − 49.9&deg; ≈ 95.1&deg;
Now we need to find c and c′.
We have
Finally,
To summarize, there are two triangles with an angle
of 35&deg;, an adjacent side of 8, and an opposite side of
6, as shown in Figure 10.13.
However, we were looking for the values for the triangle with an obtuse angle β. We can see them in the first
triangle (a) in Figure 10.13.
Solving for the Unknown Sides and Angles of a SSA Triangle
In the triangle shown in Figure 10.14, solve for the unknown side and angles. Round your answers to the nearest tenth.
sin πΎ sin π½
=
π
π
sin 85π
sin π½
=
12
9
12 π πππ½ =9 sin 85π
πΎ
Solution
In choosing the pair of ratios from the Law of Sines to use, look at the
information given. In this case, we know the angle γ = 85&deg;, and its
corresponding side c = 12, and we know side b = 9. We will use this
proportion to solve for β.
9 sin 85π
sin π½ =
12
sin π½ = 0.7471
π ππ−1 0.7471
π½ ≈ 48.3π
To find angle πΌ , apply the angle sum theorem
πΌ + π½ + πΎ = 180π
πΌ + 48.3π + 85π = 180π
πΌ = 180π − 48.3π − 85π
πΌ ≈ 46.7π
Now, only side a is needed. Use the Law of Sines to solve for a by one of the proportions.
sin πΎ π πππΌ
=
π
π
sin 85π
π ππ 46.7π
=
12
π
π sin 85π =12 π ππ 46.7π
The complete set of solutions for the given triangle is
12 π ππ 46.7π
π=
sin 85π
π ≈ 8.8
Finding the Triangles That Meet the Given Criteria
Find all possible triangles if one side has length 4 opposite an angle of 50&deg;, and a second side has length 10.
Solution
Using the given information, we can solve for the angle opposite the side of length 10. See Figure 10.15.
We can stop here without finding the value of α. Because the range of the sine function is [−1,
1], it is impossible for the sine value to be 1.915. In fact, inputting sin−1 (1.915) in a graphing
calculator generates an ERROR DOMAIN. Therefore, no triangles can be drawn with the
provided dimensions.
Solving Triangles for the Ambiguous Case (SSA)
Example
(No Triangles)
Given A = 42π , a = 3, b = 8
Since A = 42π &lt; 90π and a &lt; b, we calculate the value of sin B using the Law of Sines:
sin π΄ sin π΅
=
π
π
sin 42π sin π΅
=
3
8
3 π πππ΅ 8 sin 42π
=
3
3
8 sin 42π
sin π΅ =
3
sin π΅ = 1.784
(recall that -1 &lt; sin B &lt; 1. Hence, there are no possible triangles and nothing to solve for.
Finding the Area of an Oblique Triangle Using the Sine Function
Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the area of an oblique triangle. Recall that the area formula for a
1
triangle is given as Area =2 b h, where b is base and h is height. For oblique triangles, we must find h before we can use the area formula. Observing the two triangles in Figure
10.16, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric property sin α = opposite
β
hypotenuse to write an equation for area in oblique triangles. In the acute triangle, we have sin α = π
Or c sin α = h. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b to form a right triangle. The angle used in calculation is α′,
or 180 − α.
Thus,
Similarly,
Finding the Area of an Oblique Triangle
Find the area of a triangle with sides a = 90, b = 52, and angle γ = 102&deg;. Round the area to the nearest
integer
Solution
Using the formula, we have
Area=
1
2
Area=
1
2
ππ sin πΎ
90 52 π ππ 102π
Area ≈ 2289 π ππ’πππ π’πππ‘π
Find the area of the triangle given β = 42&deg;, a = 7.2 ft, c = 3.4 ft. Round the area to the nearest tenth.
Solution
Using the formula, we have
Area=
1
2
Area=
1
2
ππ sin π½
7.2 ππ‘ 3.4ππ‘ π ππ 42π
Area ≈ 8.19 ππ‘ 2
Solving Applied Problems Using the Law of Sines
Finding an Altitude
Find the altitude of the aircraft , shown in Figure 10.17. Round the altitude to the nearest tenth of a mile.
Solution
To find the elevation of the aircraft, we first find the distance from one station to the aircraft,
such as the side a, and then use right triangle relationships to find the height of the aircraft, h.
Because the angles in the triangle add up to 180 degrees, the unknown angle must be
180&deg;−15&deg;−35&deg;=130&deg;. This angle is opposite the side of length 20, allowing us to set up a Law
of Sines relationship.
sin 130π
sin 35π
=
20
π
20 sin 35π
π=
sin 130π
π sin 130π = 20 sin 35π
π ≈ 14.98
The distance from one station to the aircraft is about 14.98 miles.
Now that we know a, we can use right triangle relationships to solve for h.
sin
πππππ ππ‘π
=
βπ¦πππ‘πππ’π π
15π
sin 15π =
π
sin 15
β
π
β
=
14.98
β = 14.98 sin 15π
β ≈ 3.88
The aircraft is at an altitude of approximately 3.9 miles.
Example
Tamara leaves her apartment and drives west for 6.8 miles to meet her friend Gary. Leaving Gary's house, they take a road that forms a 132 degree angle with the road they were on to
drive to get ice cream. From the ice cream shop, they can take another road that leads back to Tamara's house that forms a 28 degree angle with the road they were on. How far did they
drive from Gary's to get ice cream and how far did they drive from the ice cream shop back to Tamara's? Round answers to two decimal places.
Step 1: Draw a picture representing the situation in the word problem. Label all known sides and angles.
We need to draw a triangle to represent the situation. We can fill in all three angles, since we know two of the angles and that the angles of a triangle must add up to 180 degrees.
Step 2: Set up an equation using the law of sines to solve for the unknown value. The law of sines says
sin π΄ sin π΅ sin πΆ
=
=
π
π
π
where A, B, C are the angles of a triangle and a, b, c are the sides opposite those angles, respectively.
We have two unknown values, and so will need two equations. Both equations will use the 6.8 miles and the 28π angle, since we have all the information for that angle and the side opposite it.
sin 28π
π ππ20π
=
6.8
π₯
sin 28π
π ππ132π
=
6.8
π¦
Step 3: Solve the equation from step 2 using cross-multiplication. You may need to use a calculator to finish solving and round to the specified number of decimal places if
values in the problem are not on the unit circle.
For the first equation, we can cross multiply and solve for x
sin 28π
π ππ20π
=
6.8
π₯
sin 28π π₯ = 6.8 sin 20π
For the second equation, we can cross multiply and solve for
y
sin 28π
π ππ132π
=
6.8
π¦
sin 28π π¦ = 6.8 sin 132π
6.8 sin 132π
sin 28π
6.8 sin 20π
π₯=
sin 28π
y=
x≈4.95 miles
y≈ 10.76 miles.
They drove about 4.95 miles from Gary's to the ice cream shop and about 10.76 miles from the ice cream shop back to Tamara's.
Law of Cosines
Learning Objectives
1. Use the Law of Cosines to solve oblique triangles
2. Solve applied problems using the Law of Cosines
Using the Law of Cosines to Solve Oblique Triangles
The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right
triangle ABC is placed in the coordinate plane with vertex A at the origin, side c drawn along the x-axis, and vertex C located at some point (x, y) in the plane, as illustrated in Figure
10.38. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.
We can drop a perpendicular from C to the x-axis (this is the altitude or height). Recalling the basic
trigonometric identities,
we know that
In terms of θ, x = b cos θ and y = b sin θ. The (x, y) point located at
C has coordinates (b cos θ, b sin θ). Using the side (x − c) as one leg
of a right triangle and y as the second leg, we can find the length of
hypotenuse a using the Pythagorean Theorem. Thus,
π2 = π₯ − π
2
+ π¦2
= π πππ π − π
2
+ π sin π
2
Substitute (b cos θ) for x and (b sin θ) for y.
= π2 πππ  2 π − 2 ππ πππ π + π 2 + π2 π ππ2 π
Expand the perfect square.
= π2 πππ  2 π + π ππ2 π + π 2 − 2 ππ πππ π
Group terms noting that πππ  2 θ + π ππ2 θ = 1.
π2 = π2 + π 2 − 2ππ πππ π
Factor out π2
Law of Cosines
The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the
cosine of the included angle. For triangles labeled as in Figure 10.39, with angles α, β, and γ, and opposite corresponding sides a, b, and c, respectively, the Law of Cosines is given
as three equations.
When solving for an angle, the corresponding opposite side measure is needed. We can use another version of the
Law of Cosines to solve for an angle.
Finding the Unknown Side and Angles of a SAS Triangle
Find the unknown side and angles of the triangle in Figure 10.40.
Solution
Find b
SAS
π2 = π2 + π 2 − 2accos β
π2 = 102 + 122 − 2(10)(12) cos 30π
π2 = 100 + 144− 240
3
2
π2 = 100 + 144− 120 3
π2 = 244 − 120 3
π ≈ 6.013
Find angle πΆ
π ≈ 6.013
Use cos πΌ =
π2 +π 2 −π2
2ππ
56.3π
(6.013)2 + 12 2 − (10)2
cos πΌ =
2(6.013)(12)
cos πΌ =
cos πΌ =
36.156169 + 144 − 100
144.312
80.156169
144.312
cos πΌ = 0.5554366165
πππ  −1 0.5554366165
πΌ ≈ 56.3π
Find angle πΈ
γ = 180&deg; − 30&deg; − 56.3&deg; ≈ 93.7&deg;
The complete set of angles and sides is
Solving for an Angle of a SSS Triangle
Find the angle α for the given triangle if side a = 20, side b = 25, and side c = 18
Solution
Find angle πΌ
cos πΌ =
π2 +π 2 −π2
2ππ
cos πΌ =
(25)2 +(18)2 −(20)2
2 25 (18)
625+324−400
cos πΌ =
900
549
cos πΌ =
900
cos πΌ = 0.61
πππ  −1
0.61
πΌ ≈ 52.4&deg;
Find angle π½
cos β =
π2 +π 2 −π2
2ππ
cos β =
(20)2 +(18)2 −(25)2
2 20 (18)
cos β =
400+324−625
720
cos β =
99
720
cos β = 0.1375
πππ  −1 0.1375
π½ ≈ 82.1&deg;
Find angle πΈ
πΌ + π½ + πΎ = 1800
52.4π + 82.12 + πΎ = 1800
πΎ = 1800 − 134.5π
πΎ ≈ 45.5π
Solving Applied Problems Using the Law of Cosines
Using the Law of Cosines to Solve a Communication Problem
On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the
distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to
west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower.
Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.
Solution
Using the Law of Cosines, we can solve for the angle θ. Remember that the Law of Cosines uses
the square of one side to find the cosine of the opposite angle. For this example, let a = 2420, b =
5050, and c = 6000.
=a
b=
Thus, θ corresponds to the opposite side a = 2420.
To find π π’π π π‘βπ πππ€ ππ πππ ππππ
c=
π2 = π2 + π 2 − 2ππ cos π
π2 = π2 + π 2 − 2ππ cos π
(2420ππ‘)2 = (5050ππ‘)2 + (6000ππ‘)2 − 2(5050ππ‘)(6000ππ‘) cos π
5856400ππ‘ 2 = 25502500ππ‘ 2 + 36000000ππ‘ 2 − 60600000ππ‘ 2 cos π
5856400ππ‘ 2 = 61502500ππ‘ 2 − 60600000 ππ‘ 2 cos π
5856400ππ‘ 2 - 61502500ππ‘ 2 = − 60600000ππ‘ 2 cos π
- 55646100ππ‘ 2 = − 60600000ππ‘ 2 cos π
−55646100ππ‘ 2
− 60600000ππ‘ 2 cos π
=
−60600000ππ‘ 2
−60600000ππ‘ 2
0.9182524752= cos π
πππ  −1 0.9182524752
π ≈ 23.3&deg;
To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the
position of the cell phone, as in Figure 10.43. This forms two right triangles, although we only need the right triangle that includes the first
tower for this problem.
Using the angle θ = 23.3&deg; and the basic trigonometric identities, we can find the solutions. Thus
The cell phone is approximately 4638 feet east and 1998 feet north of the first tower, and 1998
feet from the highway.
Calculating Distance Traveled Using a SAS Triangle
Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat?
Solution
The boat turned 20 degrees, so the obtuse angle of the non-right triangle is the supplemental angle, 180&deg; − 20&deg; =
160&deg;. With this, we can utilize the Law of Cosines to find the missing side of the obtuse triangle—the distance of the
boat to the port.
The boat is about 17.7 miles from port.
References:
Algebra and trigonometry LRpdf
https://study.com/
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