Dr Martin Luther King’s letter from Birmingham jail used all angles of the rhetoric triangle to address the civil rights movement in Alabama and the rest of the US. His audience was not just the clergyman that has accused of being an extremist and outside agitator, but a broader audience. He focused mainly on the moderate whites who seem to be the majority during that time and who was on either side of the fence. He tried to rationalize with them and showing examples of the issues that the blacks were facing during the civil war and how he uses his nonviolent campaign to try to put his point across and get the laws changed. He also uses simple day to day facts pointing out how segregation affects not just the blacks, but the whites also. For example, trying to explain to your young black child why they cannot play at the same park as the white kids and the effect it has on their confidence and morality.

Dr Martin Luther King’s response to the Alabama Clergyman was not just a geared to them but to a broader audience especially the moderate whites. His letter was written to reach all citizens in ever race and age group, and he did this by using the references from biblical times and other historic events to compare the civil rights movement to rationalize and offer facts and testimonies to validate his fight. For example, he used an example of the prophets from the eight century B.C. leaving their villages and carrying their “thus said the lord” far beyond their home towns to compare to his leaving Atlanta, Georgia and Birmingham, Alabama to lead the movement. Theses biblical references and connections was intentionally directed towards the clergy who had called him an extremist and an outsider; who Dr king believe would be the first t take stands on...

...
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...

...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...