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Trigonometry
Section 3 – Solve Application Problems using Right Triangle
Trigonometry
Applications
Measuring inaccessible lengths Height of a building (tree, tower, etc.) Width of a river (canyon, etc.)
Terminology
Angle of Elevation
A
Terminology
Angle of Depression
A
Application: Height
32
120 ft
h = ?
Example 1 of 4
H = 74.98 ft
To establish the height of a building, a person walks 120 ft away from the building.
At that point an angle of elevation of 32 is formed when looking at the top of the building.
Application: Height
68
h = ?
55 ft
Example 2 of 4
H = 136.1 ft
An observer on top of a hill measures an angle of depression of 68 when looking at a truck parked in the valley below.
If the truck is 55 ft from the base of the hill, how high is the hill?
Surveying
Application: Surveying
?
70 ft
37
Example 3 of 4
D = 52.7 ft
Application: Surveying
Road has a grade of 5.5%. Convert this to an angle expressed in
degrees.
100 ft
5.5 ft?
Example 4 of 4
A = 3.1
Practice Set 17
Pages 59-61
Trigonometry - Section 3
Solving problems with no right triangles.
Review
Example 1
Determine the height of this isosceles triangle.
40 15 ft
height = ?
h = 6.3 ft
Example 2
Determine the length of side x in this equilateral triangle.
height = 48”
x
x = 55.4”
Practice Set 18
Page 64 - 65
Trigonometry – Section 3
Additional Technical Applications
Application 1
Determine the depth d of the groove machined in this steel block.
82d
3”
1.1” 1.1”
d = 0.46”
0.8”
41
0.4”
41d
0.4”
Application 2
Determine the total length of steel needed to make this frame.
11 ft
35 35
h = 3.85 ft,
Total = 11 ft + 6.7 ft + 6.7 ft + 3.85 ft = 28.25 ft
Application 3
Determine the taper angle of this steel shaft.
t
145 mm
40 mm22 mm
A
t = 7.1
Application 4
The diagram shows a bolt circle. Determine the distance x between the centers of any two bolt hole locations.
+
+
+
+
x
radius 2.4”
Application 4
+
+
+
+
x
radius 2.4”
x =4.16”
Terminology: Tangent
Tangent Line
tangent line
tangent point
Property
+
90radius
+
+
Angle outside a circle
+
34 17
Putting it all together
+
40 20
1.4 ft dia.
0.7 ft radius
0.7 ft20
Example: Illustration
+
64
0.8” dia.
0.4”
32
Example: Illustration
36
0.5” dia.
+0.25”18
Example: Solve
A gauge pin is placed in a machined groove as shown. Determine the length of dimension x.
+32
8 mm dia.
x
Example: Solve
+8 mm dia.
32
x
x = 6.4 mm + 4 mm = 10.4 mm
Piston Travel
3.5”290
Piston Travel
2903.5”
1.75”70
0.599”1.75” 70
1.75” – 0.599” = 1.151”
Practice Set 19
Pages 73-75
What’s Next?
Quizzes 1 -3 pages 76 – 83
Chapter Test on Trigonometry
