Rhetorical Triangle is the active correlation between the speaker, the audience, and the situation of a presentation or speech. The rhetorical triangle determines the success of a presentation or speech. When a speech executes the three primary elements of the speaker, the audience, and the situation in balance the Rhetorical Triangle is effectively complete. It is important for a speaker to give equal relevance to all three elements to have a successful result. The Rhetorical Triangle has three equal elements that create the triangular depiction. The speaker element involves the person giving “an oral message to the listener” (Lucas, 2008, p. 17). A speaker must believe in the message in order to convince the listeners of what is said. The speaker’s state of mind at the time of the message will translate to the spectators. The speaker must also have “personal credibility” in order to gain respect from the viewers. The characteristics of the speaker also affect the audience and the situation. The speaker’s enthusiasm, deliverance, proficiency, and predispositions directly affect the ways in which the viewers will perceive the message and influence the environment. The audience element involves the person or people who will view the presentation or speech delivered by the speaker. It is important for the speaker to consider the audience element relevant so that the receivers of the message will not feel disconnected from the message. A speaker must develop a relationship with the audience. To successfully create a relationship, the speaker must research who the audience is in regard to the motivating factors, enjoyments, and biases. Each audience is different so the speaker must adjust the style of the message with each unique audience so that the listeners can effectively relate to the message. It is important for the speaker to understand that each individual in an...

...given non - collinear points is:
[Marks:1]
A.
Two
B.
infinitely many
C.
Four
D.
Three
6]
The area of a right triangle with base 5 m and altitude 12 m is
[Marks:1]
A.
50 m2
B.
15 m2
C.
9 m2
D.
30 m2
7]
Evaluate: 53 - 23 - 33
[Marks:1]
A.
80
B.
60
C.
120
D.
90
8]
The area of an equilateral triangle of side 14 cm is
[Marks:1]
A.
B.
C.
D.
9]
Simplify:
[Marks:2]...

...PLAN WEEK 5
Dr. Tonjes September 2011
LESSON: 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...

...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...

...The Mathematics 11 Competency Test
Solving 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...

...Circles
If a tangent and a secant (or chord) intersect on a circle at the point of tangency, then the measure of the angle formed 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...

...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...

...angles is :
(A) 1 and 2 (C) 2 and 3
(B) 1 and 4 (D) 4 and 5
Sample Question 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...

...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...

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