Download Math 1330 Section 6.2 1 Section 7.1: Right

Math 1330
Section 6.2
Section 7.1: Right-Triangle Applications
In this section, we’ll solve right triangles. In some problems you will be asked to find one or two
specific pieces of information, but often you’ll be asked to “solve the triangle,” which means that
you will find all lengths and measures that were not given.
You’ll use the six trigonometric functions of an angle to do this. In some cases, you will be able to
use properties of the 30° − 60° − 90° triangles or of 45° − 45° − 90° .
Example 1:
Calculate x and y from the figure below.
y
x
100cm
30o
Example 2: In ∆ABC with right angle C, ∠A = 46° and AC = 12. Find BC. Round to the answer
to the nearest hundredth.
Example 3:
Find x and y in the triangle below.
y
10 ft.
40°
x
1
Math 1330
Section 6.2
Example 4: Draw a diagram to represent the given situation. Then find the indicated measures to
the nearest degree.
An isosceles triangle has sides measuring 20 inches, 54 inches and 54 inches. What are the
measures of the angle?
Angle of Elevation
The angle of elevation is an angle that s formed by the horizontal ray and another ray above the
horizontal. So when viewing an object at a point above the horizontal, the angle between the line of
sight and horizontal is angle of elevation in the figure below.
Example 5: Draw a diagram to represent the given situation. The find the indicate measure to the
nearest tenth.
The angle elevation to the top of a building from a point on the ground 125 feet away from the
building is 8° . How tall is the building?
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Math 1330
Section 6.2
Example 6: Draw a diagram to represent the given situation. The find the indicate measure to the
nearest tenth.
A 16-foot ladder leans against a building. The ladder forms an angle of 70° with the ground.
a) How high up the building does the ladder reach?
b) What is the horizontal distance from the foot of the ladder to the base of the building?
Angle of Depression
The angle of depression is an angle that s formed by the horizontal ray and another ray below the
horizontal. So when viewing an object at a point above the horizontal, the angle between the line of
sight and horizontal is angle of elevation in the figure below.
Example 7: Draw a diagram to represent the given situation. The find the indicate measure to the
nearest tenth.
Dave is at the top of a hill. He looks down and spots his car at a 61° angle of depression. If the hill
is 59 meters high, how far is his car from the base of the hill?
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Math 1330
Section 6.2
Example 8: Mike stands 450 feet from the base of the Empire State Building and sights the top of
the building. If the Empire State Building is 1,453 feet tall, approximate the angle of elevation from
Mike’s perspective as he sights the top of the building. (Disregard Mike’s height in your
calculations.)
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Math 1330
Section 7.2
7.2 Area of a Triangle
In this section, we’ll use a familiar formula and a new formula to find the area of a triangle.
Given
a
b
1
Recall the area of a triangle is given by A = bh . We normally used this formula when we knew
2
the height of the triangle. However if the height is not given we must solve for it some way. So we
solve for the height of the triangle by breaking the given triangle down into right triangles and use
our trigonometric function (sine) to solve for the height.
Area of a Trangle
A=
1
ab sin θ
2
a, b are the lengths of two sides of a triangle θ is the angle between them.
Example 1: Find the exact area of the triangle.
30°
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Math 1330
Section 7.2
Example 2: Find the exact area of the triangle.
29
120°
17
Example 3: Find the area of an isosceles triangle with legs measuring 12 inches and the base angle
is 30 degrees each.
Example 4: If the area of ∆EFG is 30 in2, e = 8 in. and g = 15 in, find all possible measures of
angle F.
Math 1330
Section 7.2
Example 5: A regular hexagon is inscribed in a circle of radius 6m. Find the area of the hexagon.
Area of a Segment of a Circle
You can also find the area of a segment of a circle. The shaded area of the picture is an example of a
segment of a circle.
B
O
A
To find the area of a segment, find the area of the sector with central angle θ and radius OA. Then find
the area of ∆OAB . Then subtract the area of the triangle from the area of the sector.
1 2
rθ
2
Area of the segment = Areasec tor − Areatriangle
Area of a sector of a circle =
Example 6: Find the area of the segment of the circle with radius 6 meters and central angle
3π
measuring
.
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Math 1330
Section 7.3
Section 7.3: Law of Sines and Law of Cosines
Sometimes you will need to solve a triangle that is not a right triangle. This type of triangle is called an oblique
triangle. To solve an oblique triangle you will not be able to use right triangle trigonometry. Instead, you will
use the Law of Sines and/or the Law of Cosines.
You will typically be given three parts of the triangle and you will be asked to find the other three.
The approach you will take to the problem will depend on the information that is given.
If you are given SSS (the lengths of all three sides) or SAS (the lengths of two sides and the measure of the
included angle), you will use the Law of Cosines to solve the triangle.
If you are given SAA (the measures of two angles and one side) or SSA (the measures of two sides and the
measure of an angle that is not the included angle), you will use the Law of Sines to solve the triangle.
Recall from your geometry course that SSA does not necessarily determine a triangle. We will need to take
special care when this is the given information. Here’s the Law of Cosines. In any triangle ABC,
=
=
=
+
+
+
−2
−2
−2
cos
cos
cos
The development of one case of this formula is given in detail in the online text. Here’s the Law of Sines. In any
triangle ABC,
sin
sin
=
sin
=
The development of this formula is given in detail in the online text.
Here are some facts about solving triangles that may be helpful in this section.
If you are given SSS, SAS or SAA, the information determines a unique triangle. If you are given SSA, the
information given may determine 0, 1 or 2 triangles. If this is the information you are given, you will have some
additional work to do.
Since you will have three pieces of information to find when solving a triangle, it is possible for you to use both
the Law of Sines and the Law of Cosines in the same problem.
When drawing a triangle, the measure of the largest angle is opposite the longest side; the measure of the
middle-sized angle is opposite the middle-sized side; and the measure of the smallest angle is opposite the
shortest side.
Suppose a, b and c are suggested to be the lengths of the three sides of a triangle. Suppose that c is the biggest
of the three measures. In order for a, b and c to form a triangle, this inequality must be true: a + b > c . So, the
sum of the two smaller sides must be greater than the third side.
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Math 1330
Section 7.3
An obtuse triangle is a triangle which has one angle that is greater than 90°. An acute triangle is a triangle in
which all three angles measure less than 90°.
If you are given the lengths of the three sides of a triangle, where c > a and c > b, you can determine if the
triangle is obtuse or acute using the following:
If
If
, the triangle is an acute triangle.
the triangle is an obtuse triangle.
Your first task will be to analyze the given information to determine which formula to use. You should sketch
the triangle and label it with the given information to help you see what you need to find. If you have a choice,
it is
Example 1:
Find x.
x
8cm
120 
7cm
Example 2:
Find A.
7ft
5ft
A
8ft
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Math 1330
Example 3:
50cm
Section 7.3
Find x.
x
135
Example 4:
30
Find x.
60
x

75 
20in
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Math 1330
Section 7.3
Example5: Find B.
B
8m
45
4 2m
Example 6: For triangle ABC, with a  2ft , b  10ft and A= 30 , find the length of the other side and the
measure of the remaining angles.
Example 7: Triangle ABC has angle A = 40 , a = 54 m and b = 62 m. Solve for the remaining sides and
angles.
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Math 1330
Section 7.3
Example 8: Two sailboats leave the same dock together traveling on courses that have an angle of 135
between them. If each sailboat has traveled 3 miles, how far apart are the sailboats from each other?
Example 9: In ABC , B  60 and a = 17 and c = 12. Find the length of AC
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Math 1330
Section 8.1
Section 8.1: The Parabola
We already know that a parabola is the graph of a quadratic function
. But there
is more to be learned about parabolas. For example, when we studied quadratic functions, we saw that
the graphs of the functions could open up or down. As we look at conic sections, we’ll see that the
graphs of these second degree equations can also open left or right. So, not every parabola we’ll look at
in this section will be a function.
A parabola is the set of all points equally distant from a fixed line and a fixed point not on the line.
The fixed line is called the directrix. The fixed point is called the focus.
The axis, or axis of symmetry, runs through the focus and is perpendicular to the directrix.
The vertex is the point halfway between the focus and the directrix.
Basic “Vertical” Parabola:
4
Equation:
Focus: 0,
Directrix:
Focal Width: |4 |
p
Coordinates of Focal Chord:
2 ,
p
Basic “Horizontal” Parabola:
4
Equation:
Focus:
,0
Directrix:
Focal Width: |4 |
-p
p
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Math 1330
Section 8.1
Graphing parabolas with vertex at the origin:

When you have an equation, look for

If it has

Rearrange to look like

Determine p.

Determine the direction it opens.
o If p is positive, it opens right or up.
o If p is negative, it opens left or down.

Starting at the origin, place the focus p units to the inside of the parabola. Place the directrix
p units to the outside of the parabola.

Use the focal width 4p (2p on each side) to make the parabola the correct width at the
focus.
or
, it’s a “vertical” parabola. If it has
Example 1: Graph
4
or
4
, it’s a “horizontal” parabola.
. In other words, isolate the squared variable.
20 .
Vertex: _____________
Focus: ______________
Focal Width: _________
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Math 1330
Section 8.1
Example 2: 6
24
0.
Vertex: _____________
Focus: ______________
Focal Width: _________
Graphing parabolas with vertex not at the
origin:




Rearrange (complete the square) to look like
4
Vertex is , Draw it the same way, except start at this vertex.
Focus , 0 changes to
, .
Directrix:
changes to the line .



Rearrange (complete the square) to look like
Focus 0 changes to ,
.
Directrix:
changes to the line 4
.
.
.
Example 3: Find the standard form of the equation, the vertex, the focus, and the directrix for each
example:
a.
2
8
25
0
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Math 1330
b.
4
Section 8.1
2
8 0
Example 4: Suppose you know that the vertex of a parabola is at
Write an equation for the parabola in standard form.
3,5 and its focus is at 1,5 .
A line through a point that lies on a parabola is tangent to the parabola at the point if the line intersects
the parabola only at one point and the line is not parallel to the axis of the parabola. A tangent line to
the point
,
which lies on the parabola with the equation
will have slope
2
. We can use this information to find the equation of the tangent line.
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Math 1330
Section 8.1
Example 5: Write an equation of the line tangent to the parabola with the equation
5
3at
2
2
Example 6: Find the point(s) of intersection of the parabola and the line
2
8
6
5
5
5