Download TrigonometryUnit 7Oblique (non

Trigonometry
Unit 7
Oblique (non-right) Triangles
Name_____________________ Hr___
Developing the Law of Sines
In previous chapters, we have studied the properties of right triangles. The six basic trig
functions only work for right triangles. Therefore, we need to develop a method for
finding the missing parts of non-right triangles.
B
Refer to the triangle to the right.
1. Sketch the amplitude of the triangle and label it
𝑘.
2. Notice that the amplitude creates two right triangles.
Using right triangle ratios, write two statements, one
involving sin 𝐴 and one involving sin 𝐶 .
sin 𝐴 =
sin 𝐶 =
C
A
3. Notice that each of the equations in (2) involve 𝑘 . Explain why.
4. Solve each of the equations in part (2) for 𝑘 .
5. Since both the equations in part (4) are equal to 𝑘 , set them equal to each other.
6. Re-write the equation in part (5) by grouping 𝑎 with sin 𝐴 and 𝑏 with sin 𝐵.
7. Congratulations, you just discovered the Law of Sines. What information do think
is necessary to use this Law?
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Trigonometry
Unit 7
Oblique (non-right) Triangles
TRIANGLE CUT-OUT ACTIVITY
1. Take your right triangle and fold over one of the sides.
2. Cut along the side you folded to make your triangle oblique (non-right).
3. Using a RED marker, pick an angle and label it as angle A. Label the side
opposite as 𝑎. Trace a red line along the edge of side 𝑎.
4. Using a BLUE marker, follow the same procedure with angle B.
5. Using a GREEN marker, follow the same procedure with angle C.
6. Turn your triangle over and color code your sides and angles in the same way.
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Trigonometry
Unit 7
Oblique (non-right) Triangles
Using the Law of Sines
Now, we will learn how to apply the Law of Sines. The measures of the three sides and the
three angles of a triangle can be found if we know at least one side and any other two
measures. Here are the four possible cases (from Geometry):
SAA (or ASA), SSA, SAS, and SSS
Today, we will solve triangles involving the first case (SAA and ASA).
Law of Sines:
Examples: Solve the triangle.
C
1. 𝐴 = 32°, 𝐵 = 81.8°, and 𝑎 = 42.9 cm
A
B
2. 𝐴 = 35.3°, 𝐵 = 52.8°, 𝐴𝐶 = 675 ft
3. 𝐶 = 100°, 𝐴 = 30.2° , 𝑏 = 340 ft
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Trigonometry
Unit 7
Oblique (non-right) Triangles
Law of Sines (Ambiguous Case)
Today, we will solve triangles involving SSA. In this case, there may be zero, one, or two
solutions.
Solve the following triangles:
EXAMPLE 1: 𝐵 = 55°40′ , 𝑏 = 8.94 m, and 𝑎 = 25.1 m
EXAMPLE 2: 𝐴 = 55.3°, 𝑎 = 22.8 ft, and 𝑏 = 24.9 ft
EXAMPLE 3: 𝐴 = 43.5°, 𝑎 = 10.7 in. , and 𝑐 = 7.2 in.
EXAMPLE 4: Does this triangle exist? 𝐴 = 104°, 𝑎 = 26.8 m, and 𝑏 = 31.3 m
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Trigonometry
Unit 7
Oblique (non-right) Triangles
Developing the Law of Cosines
For which cases does the Law of Sines work?
Since the Law of Sines only works for certain cases, we need to develop another method to
address the other cases.
C
Refer to the triangle and follow the directions.
b
A
1. The altitude k separates triangle ABC into two right triangles.
a
k
c-x
x
B
c
Use the Pythagorean Theorem to write two equations, one relating k, b, and c-x and
the other relating a, k, and x.
2. Solve each equation for 𝑘 2 .
3. Set the equations each to other.
4. Notice that the equation in (3) involves x, however, x is not a side of triangle ABC.
Therefore, we want to write an equation that does not involve x. Expand (FOIL)
the quantity (𝑐 − 𝑥)2.
5. Solve the equation for 𝑏 2 .
6. Now we need to eliminate x from our equation. We need to substitute an equivalent
expression for x. Write an equation involving both cos 𝐵 and 𝑥.
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Trigonometry
Unit 7
Oblique (non-right) Triangles
7. Solve the equation for 𝑥.
8. Substitute the equation (7) for 𝑥 in equation (5). The resulting equation should only
involve the sides and angles from triangle ABC. This equation is called the Law of
Cosines .
9. Using a similar method, we could write two other forms of the Law of Cosines.
Based on your equation in (8), try to write the other two forms of the Law of
Cosines.
𝑎2 =
𝑏2 =
10. Now write these three forms of the Law of Cosines on the back side of your
colored triangle.
11. Answer the following questions before your leave today.
Triangle 1:
Triangle 2:
a) Can the Law of Sines be used to determine the measures of the missing angles
and/or sides? Why or why not?
b) Solve the triangles for the missing sides/angles.
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Trigonometry
Unit 7
Oblique (non-right) Triangles
Law of Cosines
In any triangle ABC, with sides a, b, and c,
𝑎2 =
𝑏2 =
𝑐2 =
Example 1: Solve triangle ABC if 𝐴 = 42.3°, 𝑏 = 12.9 m, and 𝑐 = 15.4 m.
Example 2: Solve the triangle ABC if 𝑎 = 9.47 ft, 𝑏 = 15.9 ft, 𝑐 = 21.1 ft.
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Trigonometry
Unit 7
Oblique (non-right) Triangles
Areas of Triangles
B
a
h
c
C
A
b
Remember from Geometry that the area (A) of a triangle is equal to
𝟏
𝒃𝒉.
𝟐
From the
figure, we can see that 𝒉 = 𝒂 𝒔𝒊𝒏 𝑪. We can then derive the following formulas:
𝐴=
1
𝑎𝑐 𝑠𝑖𝑛 𝐵
2
1
𝐴 = 𝑏𝑐 sin 𝐴
2
𝐴=
1
𝑎𝑏 sin 𝐶
2
EXAMPLE 1: 𝑏 = 3, 𝑐 = 4, 𝐴 = 30°
EXAMPLE 2: 𝐴 = 59.8°, 𝑏 = 15, 𝐶 = 53.1°
Heron’s Formula (also known as Hero’s formula and named after Heron of Alexandria) is
particularly useful in the SSS (side-side-side) case. It states:
𝑨 = √𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)
EXAMPLE 3: 𝑎 = 4, 𝑏 = 5, 𝑎𝑛𝑑 𝑐 = 7.
where
𝟏
𝟐
𝒔 = (𝒂 + 𝒃 + 𝒄)
EXAMPLE 4: 𝑎 = 2451, 𝑏 = 331, 𝑐 = 2427
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Trigonometry
Unit 7
Oblique (non-right) Triangles
PRACTICE:
Solve the triangles.
1. 𝐴 = 68.41°, 𝐵 = 54.23°, 𝑎 = 12.75 𝑓𝑡.
2. 𝐶 = 74.08°, 𝐵 = 69.38°, 𝑐 = 45.38 𝑐𝑚.
3. 𝐴 = 18.75°, 𝐵 = 51.53°, 𝑐 = 2798 𝑦𝑑.
4. Suppose we are given the three sides of a triangle. Explain why we can’t use the
Law of Sines to solve the triangle.
5. Suppose side a is twice as long as side b. Does it follow that angle A is twice as
large as angle B? Explain.
6. To find the distance AB across a river, a distance BC=354 m is laid off on one side
of the river. It is found that 𝐵 = 112°10′ and 𝐶 = 15°20′. Find AB
7. We wish to measure the distance across a river. We determine that
𝐶 = 112.9°, 𝐴 = 31.1°, and 𝑏 = 347.6 𝑓𝑡. Find the distance 𝑎 across the river.
A
B
C
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Trigonometry
Unit 7
Oblique (non-right) Triangles
Determine the number of triangles. (You don’t need to solve.)
8.
𝑎 = 50, 𝑏 = 26, 𝐴 = 95°
10. 𝑎 = 35, 𝑏 = 30, 𝐴 = 40°
9. 𝑏 = 60, 𝑎 = 82, 𝐵 = 100°
11. 𝑎 = 31, 𝑏 = 26, 𝐵 = 48°
Solve the triangles. If no triangle exists, explain why not.
12. 𝐶 = 41°20′ , 𝑏 = 25.9 𝑚, 𝑐 = 38.4 𝑚
13. 𝐴 = 29.7°, 𝑏 = 41.5 𝑓𝑡, 𝑎 = 27.2 𝑓𝑡.
14. 𝐴 = 142.13°, 𝑏 = 5.432 𝑓𝑡, 𝑎 = 7.297 𝑓𝑡.
15. 𝐶 = 82.2°, 𝑎 = 10.9 𝑘𝑚, 𝑐 = 7.62 𝑘𝑚
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Trigonometry
Unit 7
Oblique (non-right) Triangles
16. Without using the Law of Sines, explain why no triangle ABC exists satisfying
𝐴 = 103°20′ , 𝑎 = 14.6 𝑓𝑡, 𝑏 = 20.4 𝑓𝑡.
Solve the given triangles:
17.
18.
C
b
2
45°
B
2
B
A
4
3
95°
A
c
19.
C
20.
C
b
2
20°
5
6
A
5
B
8
A
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Trigonometry
Unit 7
Oblique (non-right) Triangles
21. A motorized sailboat leaves Naples, Florida, bound for Key West, 150 miles away.
Maintaining a constant speed of 15 miles per hour, but encountering crosswinds and
strong currents, the crew finds, after 4 hours, that the sailboat is off course by
20°.
a) How far is the sailboat from Key West at this time?
b) Through what angle should the sailboat turn to correct its course?
c)
How much time has been added to the trip because of this?
22. In attempting to fly from city A to city B, a distance of 330 miles, a pilot
inadvertently took a course that was 10° in error, as indicated in the figure.
a) If the aircraft maintains an average speed of 220 miles per hour and if the error in
direction is discovered after 15 minutes, through what angle should the pilot turn
to head toward city B?
b) What new average speed should the pilot maintain so that the total time of the trip
is 90 minutes?
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Trigonometry
Unit 7
Oblique (non-right) Triangles
Find the area of each triangle.
23.
24. 𝑎 = 9, 𝑏 = 6, 𝑐 = 4
C
2
45
4
°
\d
egr
ee
b
A
25. 𝑎 = 3, 𝑏 = 4, 𝐶 = 40°
26. 𝑏 = 1, 𝐶 = 60°, 𝐴 = 80°
27. 𝑎 = 3, 𝑐 = 2, 𝐵 = 110°
28. 𝑎 = 2, 𝑏 = 2, 𝑐 = 2
29. A perfect triangle is a triangle whose sides have whole number lengths and whose
area is numerically equal to its perimeter. Show that the triangle with side lengths
9, 10, and 17 is perfect.
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Trigonometry
Unit 7
Oblique (non-right) Triangles
30. The dimensions of a triangular lot are 100 feet by 50 feet by 75 feet. If the price
of the land is $3 per square foot, how much does the lot cost?
31. A painter needs to cover a triangular region 75 m X 65 m X 85 m. A can of paint
covers 75 m2 of area. How many cans of paint will be needed?
EXPANSION 1: A circle has a radius 𝑟 with center 𝑂. Find the area of the shaded region
as a function of the central angle 𝜃.
r
θ
EXPANSION 2: A baseball diamond is a square, 90 ft on each side, with home plate and
the three bases as vertices. The pitcher’s rubber is 60.5 ft from home plate. Find the
distance from the pitcher’s rubber to each of the bases.
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