Download Chapter 8

Geometry
2012-2013
Summary Notes
Right Triangles and Trigonometry
These notes are intended to be a guide and a help as you work through Chapter 8. These are not the only thing you need
to read, however. Rely more on your textbook for example questions worked out. After each section, successfully
completed, there is a quiz. Please notify me one class day in advance of when you intend to take each quiz. That way I
can prepare a quiz appropriately and have it ready for you.
Please note that there are two assignment options for each section. You will not complete both assignment 1 and
assignment 2. Rather, the teacher will assign you the set that you are to complete.
Thank you.
A right triangle has:
• 2 legs
• 1 hypotenuse, opposite the right angle
Label the legs of the right triangle and the hypotenuse of the right triangle.
Simplifying Radicals
Remember that in algebra we learned that we are able to simplify an answer that appears as a number under a radical
by finding factors of that number that are perfect squares, and factoring out the root of that perfect square.
Another example: √32 becomes √16 ∗ 2, which is √16√2 , since √16 = 4, this is simplified to 4√2
Simplify the following expressions – no decimals will be accepted.
363
•
50
200
Assignment 1 and 2: Complete practice worksheet: Simplifying Radicals and check your answers
8-1: Geometric Mean and the Pythagorean Theorem
a x
= and
x b
•
We can find the geometric mean, x, between two numbers a and b by setting up the proportion
•
solving to find that x = ab .
In a right triangle, with the altitude drawn from the right angle to the hypotenuse, the measure of the
altitude is the geometric mean between the measures of the two segments of the hypotenuse.
1
=
•
In a right triangle, with the altitude drawn from the right angle to the hypotenuse, the measure of a leg of
the triangle is the geometric mean between the measure of the hypotenuse and the measure of the
segment of the hypotenuse adjacent to the leg.
=
=
•
•
Assignment 1: #7-12, 24-33
Assignment 2: #24-37, 42, 45, 46
8-2: The Pythagorean Theorem and Its Converse
• The Pythagorean Theorem tells us that in a right triangle, the sum of the squares of the measures of the legs
equals the square of the measure of the hypotenuse. We use this idea to find the missing side of any right
triangle, if we know 2 of the sides.
+ = Find the third side of each right triangle, ABC. Be sure to give your answer as a simplified radical where applicable – no
decimals.
1. a = 4, b = 3, c = _____
•
2. c = 10, b = 8, a = ____
3. b = 6, a = 12, c = ___
The converse of the Pythagorean Theorem tells us “If the sum of the squares of the measures of two sides of
a triangle equals the square of the measure of the longest side, then the triangle is a right triangle.” We use
this idea to decide whether three measures can possibly create a right triangle.
Tell whether each set of measures can make a right triangle. Show your work.
1. 30, 40, 50
2. 14, 23, 35
2
3. 12, 5, 13
•
A Pythagorean triple is a group of three whole numbers that satisfies the equation a 2 + b 2 = c 2 where c is
the greatest number. If the three numbers satisfy the Pythagorean Theorem, and they are whole numbers,
then they are a Pythagorean triple.
•
•
Assignment 1: #4, 6-14, 19-25 odd
Assignment 2: #19, 21, 23, 27-29, 34-37
8-3: Special Right Triangles
•
2 times as long as a leg.
In a 45°-45°-90° triangle, the hypotenuse is
√21
1
2
3√2
1
2
We can use this “model” of the 45°-45°-90° triangle to write a proportion that will let us solve for any missing
side measure in any 45°-45°-90° triangle.
3√2
1 1
√2
2 2
1 1 √2
3 times
•
In a 30°-60°-90° triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is
as long as the shorter leg.
•
We use the idea of similar triangles to calculate the missing measures for 30°-60°-90° triangles, using
proportions that look like: = = because we know the relationships of the sides one to another in the
√
30°-60°-90° triangle.
6
x
2
30°
√3
g
4√3
y
60°
60°
1
•
•
6
1 √3 2
Assignment 1: #9-21 odd, 22, 32, 33-36
Assignment 2: #16-23, 32-36, 38
3
z
4√3 1
2
√3
8-4: Trigonometry
•
The word “trigonometry” comes from two Greek terms – trigon (triangle) and metron (measure). The study
of trigonometry involves triangle measurement.
•
The three trigonometric ratios are sine, cosine and tangent. These are ratios of sides of the triangle as
related to the acute angles of the right triangle. We cannot use the trigonometric ratios on right angles.
•
Use SOH CAH TOA to remember what each ratio stands for (sine equals opposite over hypotenuse,
cosine equals adjacent over hypotenuse, tangent equals opposite over adjacent)
o When we solve problems involving sin, cos, tan, we need to remember that the operation that
“undoes” each ratio is the inverse of that ratio, or sin-1, cos-1, tan-1
Answer the following, drawing right triangles that model the situation:
o A road is inclined at an angle of 10º with the horizontal. Find the distance that must be driven on the
road in order to be elevated 15 feet above the horizontal.
o Janet Owens is an architect designing a new parking garage for the city. The floors of the garage are
to be ten feet apart. The exit ramps between the floors are to be 75 feet long. What angle does the
ramp make with the horizontal?
•
•
Assignment 1: #18-23 all, 24, 25, 38-51
Assignment 2: #24, 25, 50-56
o
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8-5: Angles of Elevation and Depression
• Both angles of elevation and angles of depression are measured from the horizontal. Remember that
elevation goes from the ground up, while depression will be an angle from the upper horizontal,
downwards.
• When putting together problems in angles of depression and elevation, break the problem down into small
pieces, and draw a picture to show each piece. Build each part of the question onto the first and think about
what the real-life situation looks like.
• Use the diagram you drew, on which you wrote the values that you know and what you are looking for, to
decide which trigonometric function you want to use to solve for the unknown.
Therese is flying a kite whose string makes a 62° angle with the ground. The kite string is 83 meters long. How far is the
kite above the ground?
•
•
Assignment 1: #4-13 all
Assignment 2: Worksheet
8-6: The Law of Sines
• The Law of Sines (and the Law of Cosines in the next section) applies to any triangle, not just a right triangle.
Therefore, this proportion will be more useful to us, since we have many more real-life situations that are
not modeled well by right triangles.
•
Let ∆ABC be any triangle with a, b, and c representing the measures of sides opposite angles with measures
A, B, and C, respectively. Then,
•
•
•
•
sin A sin B sin C
=
=
.
a
b
c
We can use the Law of Sines to solve a triangle (find all the missing measures) when:
1. You know the measures of two angles and any side of a triangle
2. You know the measures of two sides and an angle opposite one of these sides of the triangle.
Procedure
o Draw triangle that we are concerned with.
o Label knowns and unknowns.
o Set up Law of Sines for this particular triangle (not numbers yet)
o Substitute what we know into the Law of Sines for this triangle.
o Find proportions that you can solve (need to have only one variable in each proportion).
Assignment 1: #13-17 odd, 19, 21, 27, 28, 29
Assignment 2: #23, 25, 27-33 (or worksheet)
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8-7: The Law of Cosines
• Let ∆ABC be any triangle with a, b, and c representing the measures of sides opposite angles with measures
A, B, and C, respectively. Then, the following equations hold true:
o a 2 = b 2 + c 2 − 2bc cos A
o b 2 = a 2 + c 2 − 2ac cos B
o c 2 = a 2 + b 2 = 2ab cos C
• We use the Law of Cosines to solve a triangle when:
1. You know the measures of two sides and the included angle.
2. You know the measures of three sides.
• We can use the Law of Cosines and the Law of Sines together, so that we don’t have to use the long Law of
Cosines for every piece of the problem.
•
•
Assignment 1: #16-18, 19, 21
Assignment 2: #25-31 odd, 33, 36
Solve each triangle using the Law of Sines and / or the Law of Cosines. Please place all answers for each triangle
together, and box the entire set. Round all values to the nearest tenth.
a 2 = b 2 + c 2 − 2bc cos A
Recall that: b 2 = a 2 + c 2 − 2ac cos B
c 2 = a 2 + b 2 − 2ab cos C
1. a = 19, m∠A = 39°, m∠B = 78°
2. a = 27, b = 39, c = 15
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