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...

...Similarity and congruence
Two triangles are said to be similar if every angle of one triangle has the same measure as the corresponding angle in the other triangle. The corresponding sides of similar triangles have lengths that are in the same proportion, and this property is also sufficient to establish similarity.
A few basic theorems about similar triangles:
* If two corresponding internal angles of twotriangles have the same measure, the triangles are similar.
* If two corresponding sides of two triangles are in proportion, and their included angles have the same measure, then the triangles are similar. (The included angle for any two sides of a polygon is the internal angle between those two sides.)
* If three corresponding sides of two triangles are in proportion, then the triangles are similar.
Two triangles that are congruent have exactly the same size and shape: all pairs of corresponding interior angles are equal in measure, and all pairs of corresponding sides have the same length. (This is a total of six equalities, but three are often sufficient to prove congruence.)
Some sufficient conditions for a pair of triangles to be congruent are:
* SAS Postulate: Two sides in a triangle have the same length as two sides in the other...

...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=""...

...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...

...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...