Problem 1: A ladder is leaning against the side of a house and forms a 67 degree angle with the ground. The foot of the ladder is 18 feet from the house. Find the length of the ladder.

For the first problem on the major grade E-portfolio the first step would be to label the triangle opp, adj, and hyp. Then you write the equation with the formula which should give you Cosine(67)=18/x. Then you have to solve for theta using cosine which changes the formula to Cosine(0.391)=18/x, Then you have to multiply the denominator by both sides. Then because we need to isolate the x variable we need to divide theta by both sides which would then requires us to divide 18 by 0.391 which would give the answer x=46.036.

Problem 2: Solve x by using Pythagorean theorem

First we input the leg and hypotenuse values into the formula a^2 + b^2= c^2 which should become a^2+13^=17^2. The next step is to square two given numbers 13 and 17 which should change our equation to a^2+169=289. Then we subtract 289-169 which gives the number 120. Finally to get our answer we have to find the square root of 120 which is 10.954. Which gives our answer x=10.954.

Problem 3: Solve x by using Trig Ratios

To solve for x the first thing to do is to find which formula is needed to solve the problem, since the two numbers given are in the opp and adj places on the triangle then they tell us that we need to use the cosine formula of adj/opp. The first step to solve the equation is to do Cosine(15)=22/x. We solve cosine(15) which gives us 0.966 which we then need to add into our equation. cosine(0.966)=22/x the next step is to multiply the denominator by both sides which cancels itself out on the right. 0.966x=22. The next step is to get x by itself which causes us to have to divide 0.966x by both sides which then gives us our answer x=22.774.

Problem 4: Solve x by using inverse Trig Ratios

To solve forhing that is required is for us to label our triangle adj, opp, and hyp.  After labeling we need to find out which formula is going to be required. Since we are given the opp and hyp we can determine that we are going to be using sin-1=18/25. The next step that we need to do is to divide 18 by 25 which gives us 0.621. And to do inverse trig we determine the answer by using sin-1(0.621) which then gives us the answer x=38.389.

Problem 5: Solve the Triangle

A. Solve C by using law of  Cosines

B. Solve A by using Law of Sines

CSolve B by using 180-_-_

To solve for C by using Law of cosines first we need to find out which formula we are going to need. Since the triangle gives us one angle and two sides we can determine that we are going to be using the SAS triangle formula. The formula is c^2= a^2+b^2-2abCosC. After determining which fornula that we are going to need is to plug in our numbers which would give us the equationc^2=30^2+20^2-2(30×20)xCos(59). The next step is to square our numbers which changes the equation to c^2=900+400-2(30×20)xCos(59). The next step is to multiply our numbers which would give us c^2=900+400-1,200 x Cos(59). The next step is to solve for Cos(59) which changes our equation to c^2=900+400-1200 x 0.515. Then we solve the equation by add 900+400 subtract the sum from 1,200 then multiply the answer by 0.515 which would give us c^2=682. Although since we are not solving for c^2 but instead for c we need to find the square root of 682 which gives us our answer 26.115.

To solve for A using the law of sines ,First you have to insert your numbers A a C c into the formual x/sinA=x/sinB. After inserting the numbers into the formula the formula should look like 30/sinA=26.115/sin59. The next step is to solve for sin which should then cause the equation to look like this 30/sinA=26.115/0.857. The next step is to cross multiply the two fractons which should cause the equation to look like this 25.71=26.115A. The next step is to isolate the variable which requires us to divide 26.115 by both sides which causes the equation to look like this sin-1(0.984) The final step is to solve the last little equation which gives the answer A=79.737

To find B the only thing that you have to do is to subtract 180 from A and C. 180-79.737-59=41.263.

 

Note: I am still having trouble inserting my images please do not deduct points on my MAJOR GRADE eportfolio Please

 

 

 

 

 

 

When solving for law of sines the first step would be to label the triangle so that you are able to determine which formula that you are suppose to use for it. Then the next step is to write the equation x/x=x/x. Next it is required to cross multiply the two numbers. After the numbers have been multiplied the variable s required to be on it’s own so it is divided which gives us our answer.

 

 

Note: I am still having trouble inserting an image please don’t take points off of my assignment.

I am thankful for the food that I will be able eat on Thanksgiving day as because I know that not everyone has food to eat on that day. I am thankful to be able to share this dinner with my family as because it makes me feel helpful to other people. I am thankful for my parents because they are always there when and if anything ever happens and they will always help me out whenever possible. I am thankful for my friends who are always there with me throw the bad times and the good times, they always know how to cheer me up and help me when things go bad. And finally I am thankful for all of the animals in the world there is not a lot to say why I just really like all of the cute animals in the world.

Trigonometric ratios is nothing like doing Pythagorean theorem. The formulas for doing trigonometric ratios is very easy to forget and is not something that I can just say over and over again to get it. The formula for Sin is opp/ hyp, Cosine is adj/hyp , and Tangent is opp/adj. A small tip that helps me remember this is that the acronym SOH,CAH,TOA. which is the best way to help me remember the formulas. I believe that with this acronym I may be able to do well on my test.

The Pythagorean theorem formula a^2 + b^2 = c^2 is used when trying to find the missing side of an right triangle. The formula is (in my opinion) the easiest formula to use. To solve for the side of an right triangle known as the hypotenuse which is the longest side of the triangle. To solve for the hypotenuse you take the number of the two given legs and square them to the second power. Next you add the two numbers together. Then you take the sum and find the square root of it.

 

 

 

 

(note I am having trouble inserting image please help.

On the notes for rotation in class it was taught that the formulas for counterclockwise and clockwise are simply reversed for all except the formula for 180 degrees. The formulas for counterclockwise are; 90 degrees swap the place values for the x, and y coordinate and then multiply the y values by negative one. This is how example number one on the study guide received its answer.(4,-1) – (1,4)

One the notes for linear measure the rules for finding the total distance between two points is by adding the two number together. Which is how the example for number 1 received it’s answer. Find BD: 16.8 mm plus 50.4 mm. Which adds together to get 67.2 or C.

For the distance formula I chose to do number 7 I forgot the proper way to calculate it so I had looked up the formula off of google. The Distance formula formula is the square root of (x2+x1)^2 +(y2+y1)^2. After knowing this I inserted my problem into my formula which gave me    (5-(-7)^2+(9-0)^2. The next step to do is to switch the minus sign to a plus sign so after that is completed the next step is too add the two number together which gives the square root of (12)^2 plus (9)^2. The next step is too square the two numbers which gives the problem the square root of 144 plus 81. The next step is too add which gives the problem the square root of 225 which after solving is 15 which is the solution.

For the reflection test question I chose to do question number 2. The rule for reflection over the y-axis is to multiply the x place value by -1. The points for the problem is (-3,0) (-4,5) (-7,5) (-6,0). After multiplying all of the x place values by negative 1 the solutions should be. (3,0) )4,5) (7,5) (6,0)

 

Over the course of the nine weeks the biggest struggle for me was to try to learn the formulas so that I could do well on the tests.To be more specific I would say that the distance and midpoint formulas were the hardest to learn. I personally liked the Reflection, Translation, Rotation, and Dilation assignments because I thought that they were the most fun to complete. The most interesting fact that I learned all year is the fact that distance can never be negative on matter how far you go back.

I chose to complete number 2 on the study guide. I chose number 2 because I thought that it would of been the easiest to explain. The main rule in number 2 is to multiply the y coordinate by -1 the reason that we do this is because we are attempting to reflect over the x axis. Do following our rule I have determined that the way to solve the problem is to multiply the points:(3,-4) (1,-6) (-2,-3) (3,1). So after multipying the y values by -1 the points are (3,4) (1,6) (2,3) (3,-1). After we determine our points all that is left is to graph the points.

In class when discussing reflection is referred to as flipping the image across the axis. The formula for reflection across the x axis is y=-x^2. The formula for reflection across the y axis is y, -x. The formula are easy to learn but are also easy to forget if not gone over within a week. The main secret to quickly learn how to do reflections is that the image is going to be congruent to the pre image so by that conclusion it is determined that the problem is wrong 9 times out of 10.

The formula for rotations is a little difficult for me to understand, but after a while I was able to understand it quite easily. One of the most important things to know is which way to rotate, the problem will state whether or whether not the figure is rotating clockwise or counter-clockwise. A cool little fact that helps learn the formulas is that 90° clockwise formula is the same as 120° counterclockwise. The 180° turn is the same for both no matter the way that it is being turned. It is important to pay attention which way the problem tells to turn the figure or image.

The Questions on the study guide that I chose to explain how I got my answer are Questions 4 and 12.

 

The problem for 4 is to state another name for line m because the line consists of three points we have to remove the middle point and only include the other two points on the plane so another name for line l is AC.

 

The problem for 12 is to find the measure of the line segment which can be determined by subtracting the total number of the line segment by the given line, so the line segment is 15mm and the given line is 9.7mm so the missing line is 5.3mm.