Presentation Essentials
Tamie Martz
PRES111-1301B-02
March 30 2013
American Intercontinental University

Abstract
The rhetorical triangle is used to, prepare for a speech there are three points to the rhetorical triangle, the speaker, audience, and the situation. It is also a very useful tool, to help speakers with their presentations. Speakers also needs to flexible in communicating with their audience.

Rhetorical triangle

September 11 2011, a day when tragedy struck New York City four passengers planes were high jacked by terrorist. Two planes hit the twin towers, one hit the Pentagon, and a plane crashed in PA, it was too believed to be high jacked as well. This horrible event was witnessed by millions of people of television viewers it was the worst terrorist attack on the American soil, it was also the most lethal terrorism in human experience. President Bush address the Congress of the following the attaches of 9/11. As the speaker he talks to the audience about the attacks of 9/11 he states that no report is needed, it already has been delivered by American people. President Bush also says “we have seen it in the courage of passengers who rushed a plane into the ground, passengers like exceptional man named Todd Beamer.”He talks about how our country was awakened to danger to defend our freedom. Bush expressed his concerns about what happened when we were attached by the terrorist. He also said “Prayer has comforted us in sorrow, and will help strengthen us for the journey ahead." When President Bush gave his speech about the attacks of 9/11 he focused on the months ahead, life will get back to normal. Within the Rhetorical triangle I believe President Bush gave a great speech he had the attention of the Congress and the American people. George W Bush first address of the United States Nations General Assembly President Bush stands in front a group of people in the General Assembly, talking about the event that took place on Sept. 12...

...
5C Problems involving triangles
cQ1. The diagram shows a sector AOB of a circle of radius 15 cm
and centre O.
The angle at the centre of the circle is 115.
Calculate (a) the area of the sector AOB.
(b) the area of the shaded region. (226 , 124
nQ2. Consider a triangle and two arcs of circles.
The triangle ABC is a right-angled isosceles
triangle, with AB = AC = 2.
The point P is the midpoint of [BC].
The arc BDC is part of a circle with centre A.
The arc BEC is part of a circle with centre P.
(a) Calculate the area of the segment BDCP.
(b) Calculate the area of the shaded region BECD.
cQ3. In the following diagram, O is the centre of the circle
and (AT) is the tangent to the circle at T.
If OA = 12 cm, and the circle has a radius of 6 cm,
ﬁnd the area of the shaded region.
cQ4. The diagram shows a circle, centre O, with a radius 12 cm.
The chord AB subtends at an angle of 75° at the centre.
The tangents to the circle at A and B meet at P.
(a) Using the cosine rule, show that the length of AB
is
(b) Find the length of BP.
(c) Hence find
(i) the area of triangle OBP;
(ii) the area of triangle ABP.
(d) Find the area of sector OAB.
(e) Find the area of the shaded region.
Miscellaneous Problems
Q5. The diagram below...

...Oblique Triangles, Laws of Sines and Cosines
INTRODUCTION:
Student will demonstrate how to apply laws of sines and cosines to oblique triangles.
OBJECTIVES:
After completing this unit, the student will be able to:
6. Use the Law of Sines and the Law of Cosines to solve oblique triangle problems.
6.1. Summarize the Law of Sines.
6.2. Find the area of an oblique triangle using the sine function.
6.3. Judge when an ambiguous case of the Law of Sines occurs.
6.4. Solve applied problems using the Law of Sines.
6.5. Summarize the Law of Cosines.
6.6. Use the Law of Cosines to solve oblique triangle problems.
6.7. Solve applied problems using the Law of Cosines.
6.8. Find the area of an oblique triangle using Heron’s formula.
PROCEDURE:
Content
Activity
Objectives
Present objectives and purpose of lesson.
Law of Sines and Law of Cosines
Generalize the sine and cosine relations of the right triangles to oblique triangles by defining the two laws.
Law of Sines: This law relates the three sides of any triangle to the angles opposite the sides, typically labeled a, b, and c for the sides and A, B, and C for the angles.
Law of Cosines: This law relates one side to the other two sides and its corresponding angle:
Relate that either of these relations reduce to simpler forms for the case of right triangles, particularly:
1. Law of Sines...

...Finding an Angle in a Right Angled Triangle
You can find the Angle from Any Two Sides
We can find an unknown angle in a right-angled triangle, as long as we know the lengths of two of its sides.
￼
Example
A 5ft ladder leans against a wall as shown.
What is the angle between the ladder and the wall?
(Note: we also solve this on Solving Triangles by Reflection but now we solve it in a more general way.)
The answer is to use Sine, Cosine or Tangent!
But which one to use? We have a special phrase "SOHCAHTOA" to help us, and we use it like this:
Step 1: find the names of the two sides you know
￼
Example: in our ladder example we know the length of:
the side Opposite the angle "x" (2.5 ft)
the long sloping side, called the “Hypotenuse” (5 ft)
Step 2: now use the first letters of those two sides (Opposite and Hypotenuse) and the phrase "SOHCAHTOA" to find which one of Sine, Cosine or Tangent to use:
￼￼￼
In our example that is Opposite and Hypotenuse, and that gives us “SOHcahtoa”, which tells us we need to use Sine.
Step 3: Put our values into the Sine equation:
Sin (x) = Opposite / Hypotenuse = 2.5 / 5 = 0.5
Step 4: Now solve that equation!
sin (x) = 0.5
Next (trust me for the moment) we can re-arrange that into this:
x = sin-1 (0.5)
And then get our by Text-Enhance" href=""...

...Problems with Similar Triangles
In the previous document in this series, we defined the concept of similar triangles, ∆ABC ∼ ∆A’B’C’ as a pair of triangles whose sides and angles could be put into correspondence in such a way that it is true that property (i): A = A’ and B = B’ and C = C’. property (ii):
a b c = = a' b' c '
If property (i) is true, property (ii) is guaranteed to be true. If property (ii) is true, then property (i) is guaranteed to be true. We also demonstrated some strategies for establishing that two triangles are similar using property (i). This is very useful to be able to do, since then, we may be able to use the property (ii) conditions to calculate unknown lengths in the triangles. Example 1: Given that lines DE and AB are parallel in the figure to the right, determine the value of x, the distance between points A and D. solution: First, we can demonstrate that ∆CDE ∼ ∆CAB because C=C and ∠CDE = ∠CAB because line AC acts as a transversal across the parallel lines AB and DE, and since ∠CDE and ∠CAB are corresponding angles in this case, they are equal. Since two pairs of corresponding angles are equal for the two triangles, we have demonstrated that they are similar triangles. To avoid error in exploiting the similarity of these triangles, it is useful to redraw them as separate triangles:
B 11
D 7 E
B E 11 A 7 x...

...Similar Triangles Project
February 12, 2013
Introduction:
In this project, I found the height of an object I chose based on how tall one of my partners is, how far away she is from the mirror, and how far the mirror was from the base of one of the objects. From there I set up a proportion and solved for X. X represented the unknown height of the chosen object. Once I figured this out I then converted to feet and compared that to my partners height to see if it was a reasonable or realistic height.
Two-Column Proof:
|Statements |Reasons |
|The triangles are right triangles |Given—Mr. Visser told us that we can assume this |
|Triangles are similar |If there exists a correspondence between the vertices of two |
| |triangles such that two angles of one triangle are congruent to |
| |the corresponding angles of the other, then the triangles are |
| |similar. |
Conclusion:
In...

...is half the measure of its intercepted arc.
�
Possible answer: It is given that AB � EB. So �ABE is an isosceles triangle, 1 � and �BAC � �BEA. �BEA is an_ inscribed angle, so m�BEA � __ mBC. By _ 2 1 � substitution, m�BAC � __ mBC. AD and AE are secants that intersect in the 2 1 DE __ (m� � m�). Substitution leads exterior of the circle. So m�BAC � BC 2 � � � 1 � 1 � to __ mBC � __ (mDE � mBC ). This simplifies to mDE � 2mBC . � � 2 � 2� � �
2. Given: RS � TU, RU � ST Prove: Q is the center of the circle. (Hint: Show that Q is equidistant from three points on the circle.)
� � �
_
�
Tangent BC __› and secant BA intersect at B.
� �
__ ›
1 � m�ABC � __ mAB 2 If two secants or chords intersect in the interior of a circle, then the measure of the angle formed is half the sum of the measures of its intercepted arcs.
� �
�
�
Possible answer: Draw chords RU and _ It is given that RU �� Because ST. _ ST. � congruent arcs have congruent chords, RU � ST. It is given that RS � TU. �RUS, �URT, �TSU, and �STR are all inscribed angles that intercept either � � RS or TU. Therefore all four angles have the same measure and are congruent. _ the By SAS, �QRU and �QST are congruent triangles. Furthermore, _ base _ _ angles are all the same, so they are isosceles triangles. So RQ, UQ, and SQ (and TQ ) are congruent by CPCTC and the Isosceles Triangle Theorem. Congruent segments have equal lengths, so Q is...

...Paper
351
10. In ∆ABC, the bisector ∠A is same as the median through A. ∆ABC is : (A) isosceles with AB = BC (C) isosceles with AB = AC (B) a right angled triangle (D) isosceles with BC = AC
11. The area of a circle is 314 cm2. If π = 3.14, then its diameter is : (A) 100 cm (C) 20 cm (B) 50 cm (D) 10 cm
12. The total surface area of a closed right circular cylinder of radius 3.5 m and height 7 m is : (A) 77 m2 (C) 231 m2
3 , then sin θ is : 4 3 5 5 4
(B) 154 m2 (D) 308 m2
13. If tan θ =
4 5 4 3
(A)
(B)
(C)
(D)
14. If sin θ + cos θ = 2 cos θ , the value of tan θ is : (A) (C)
2 2 −1
(B) 1 (D)
2 +1
15. Cumulative frequency of a class in a frequency distribution table is : (A) total of all frequencies (B) sum of all frequencies prior to that class (C) sum of all frequencies upto the class (D) sum of frequencies after the class
352
Mathematics
16. The cumulative frequency of the last class of a frequency distribution is equal to : (A) frequency of last class (B) frequency of first class (C) frequency of the class prior to the last class (D) total of all frequencies
17. Simplify :
3125 343
18. Expand :
Fx 2 H
2
−1 x
I K
2
19. Factorize : 1 – x4y4
p 20. Express − 0.3 in the form − q where p and q are natural numbers.
21. The sides of a triangle are in the ratio 1 : 1.5 : 2. If the perimeter is 13.5 cm, find the length of each side. 22. The entries in the pass book of...

...answer & solutions on a separate answer sheet.
A. True or False.
______1. The area of a triangle equals one-half the product of two of its side lengths and the sine of the angle.
______2. Given only the three sides of a triangle, there is insufficient information to solve the triangle.
______3. Given two sides and the included angle, the first thing to do to solve the triangle is to use the Law of Sines.
______4. The Law of Sines states that the ratio of the sine of an angle in a triangle to its opposite side is
equal to the ratios of the sines of the other two angles to their opposite sides.
______5. Law of Cosines says that the square of any side of a triangle is equal to the squares of the sum of the other two sides, minus twice the product of those two sides times the cosine of the included angle.
______6. If you are given the lengths of two sides of a right triangle, you can solve the right triangle.
______7. If you are given the length of the hypotenuse of a right triangle and the measures of the angle opposite the hypotenuse, you can solve the right triangle.
______8. The semiperimeter of any triangle is one-half the sum of its angles.
______9. A businessman wishes to buy a triangular lot in a busy downtown location. If the lot frontages on the three adjacent streets are 125, 280, and 315 ft., then area...