section 7.1 – solving right trianglesalmus/1330_chapter7.pdf · 2015-05-02 · 1 section 7.1 –...

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1 Section 7.1 – Solving Right Triangles Note that a calculator will be needed for most of the problems we will do in class. Test problems will involve angles for which no calculator is needed (e.g., 30 , 45 , 60 , 120 ° ° ° ° , etc.). So, you will still need those unit circle values. We’ll use right triangle trigonometry to find the lengths of all of the sides and the measures of all of the angles. 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,” that is, to find all lengths and measures that were not given. Example 1: Find x and y. x y 10 25 0

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Page 1: Section 7.1 – Solving Right Trianglesalmus/1330_chapter7.pdf · 2015-05-02 · 1 Section 7.1 – Solving Right Triangles Note that a calculator will be needed for most of the problems

1

Section 7.1 – Solving Right Triangles

Note that a calculator will be needed for most of the problems we will do in class. Test problems will involve angles for which no calculator is needed (e.g., 30 , 45 , 60 ,120° ° ° ° , etc.). So, you will still need those unit circle values. We’ll use right triangle trigonometry to find the lengths of all of the sides and the measures of all of the angles. 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,” that is, to find all lengths and measures that were not given. Example 1: Find x and y.

x y

10

250

Page 2: Section 7.1 – Solving Right Trianglesalmus/1330_chapter7.pdf · 2015-05-02 · 1 Section 7.1 – Solving Right Triangles Note that a calculator will be needed for most of the problems

2

Example 2: In ABC∆ with right angle C, °=∠ 40A and AC = 12. Find BC. Round the answer to the nearest hundredth.

Page 3: Section 7.1 – Solving Right Trianglesalmus/1330_chapter7.pdf · 2015-05-02 · 1 Section 7.1 – Solving Right Triangles Note that a calculator will be needed for most of the problems

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Example 3: Draw a diagram to represent the given situation. Then find the indicated measures to the nearest tenth of a degree. An isosceles triangle has sides measuring 20 inches, 54 inches and 54 inches. What are the measures of its angles?

Page 4: Section 7.1 – Solving Right Trianglesalmus/1330_chapter7.pdf · 2015-05-02 · 1 Section 7.1 – Solving Right Triangles Note that a calculator will be needed for most of the problems

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Example 4: Draw a diagram to represent the given situation. Then find the indicated measure to the nearest tenth. A 216 foot ladder is leaned against the side of a building. If the ladder forms a 41° angle with the ground, how high up the side of the building does the ladder reach?

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Angle of Elevation; Angle of Depression:

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Example 5: Draw a diagram to represent the given situation. Then find the indicated measure to the nearest tenth. The angle of 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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Example 6: Draw a diagram to represent the given situation. Then find the indicated measure to the nearest tenth. Dave is at the top of a hill. He looks down and spots his car at a 60° angle of depression. If the hill is 48 meters high, how far is his car from the base of the hill?

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Section 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.

You have probably used the formula bhK2

1= to find the area of a triangle, where b is

the length of the base of the triangle and h is the height of the triangle. We’ll use this formula in some of the examples here, but we may have to find either the base or the height using trig functions before proceeding. Here’s another approach to finding area of a triangle. Consider this triangle:

The area of the triangle ABC is: )sin(2

1AbcK =

It is helpful to think of this as Area = ½*side*side*sine of the included angle. Example 1: Find the area of the triangle.

b a

c

A

C

B

14

6

450

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Example 2: Find the area of the triangle. Example 3: Find the area of an isosceles triangle with legs measuring 12 inches and base angles measuring 52 degrees each. Round to the nearest hundredth. Example 4: In triangle ABC; 12, 20a b= = and 42.0)sin( =C . Find the area of the triangle.

120°°°°29

1712

20

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Example 5: In triangle KLM, k = 10 and m = 8. Find all possible measures of the angle L if a) the area of the triangle is 20 unit squares. b) the area of the triangle is 25 unit squares. c) the area of the triangle is 80 unit squares.

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Formula for Area of a Regular Polygon Given a Side Length

2

4 tan

S NA

N

π=

, where S = length of a side, N = number of sides.

Example 6: A regular hexagon is inscribed in a circle of radius 12. Find the area of the hexagon. For reference, a pentagon has 5 sides, a hexagon has 6 sides, a heptagon has 7 sides, an octagon has 8 sides, a nonagon has 9 sides and a decagon has 10 sides.

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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.

A

O B

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.

Area of segment = Area of sector AOB - Area of AOB∆

= )sin(2

1

2

1 22 θθ rr −

Example 7: Find the area of the segment of the circle with radius 8 inches and central

angle measuring 4

π.

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Section 7.3 - The Law of Sines and the 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. Please read this before class! 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. This is called the “ambiguous case” of the law of sines. 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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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 222 cba >+ , the triangle is an acute triangle. If ,222 cba <+ 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 usually best to find the largest angle first.

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THE LAW OF SINES AND THE LAW OF COSINES Here’s the Law of Sines. In any triangle ABC,

.sinsinsin

c

C

b

B

a

A ==

USED FOR SAA, SSA cases! SAA: One side and two angles are given SSA: Two sides and an angle opposite to one of those sides are given Example 1: Find x.

b a

c A

C

B

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Here’s the Law of Cosines. In any triangle ABC,

Cabbac

Baccab

Abccba

cos2

cos2

cos2

222

222

222

−+=

−+=−+=

USED FOR SAS , SSS cases!

SAS: Two sides and the included angle are given SSS: Three sides are given

b a

c A

C

B

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Example 2: In ,ABC∆ a = 5, b = 8, and c = 11. Find the measures of the three angles to the nearest tenth of a degree.

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Example 3: In .18 and 78,26, =°=∠°=∠∆ yZXXYZ Solve the triangle. Give exact answers.

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Example 4: In , 50 , 9 and 6.ABC A b a∆ ∠ = ° = = Solve the triangle and round all answers to the nearest hundredth.

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Example 5: 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?

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Example 6: In ,ABC∆ 60B∠ = ° , a = 17 and c = 12. Find the length of AC.

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Note: SSA case is called the ambiguous case of the law of sines. There may be two solutions, one solution, or no solutions. You should throw out the results that don’t make sense. That is, if sin A > 1 or the angles add up to more than 1800. SSA Case (Two sides and an angle opposite to those sides)

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Example 7: In .7 and 5,112, ==°=∠∆ qpPPQR How many possible triangles are there? Solve the triangle. Round the answers to three decimal places.

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Example 8: In 5,7,22, ==°=∠∆ xyYXYZ . How many possible triangles are there? Solve the triangle and round all answers to the nearest hundredth.