Alicia's Comments

okay im going to comment one of my own questions.

I have trouble with determining when to use the law of sines and law of cosines. If you have this same problem, maybe this explanation will help.

Basically, only use the law of sines when you have a non-right triangle that has a pair. If it's a right triangle, then you would use SOHCAHTOA. You also use law of sines to find the other angles when doing law of cosines. Use law of cosines for non right triangles that have no pairs.

I hoped this helped more than just myself :)

Alicia's Reflection #15

okay so im going to go over some older concepts from the beginning of trigonometry since our midterm is accumulative.

Unit Circle:

90 degs. = (0,1) pi/2

180 degs. = (-1,0) pi2

70 degs. = (0,-1) 3pi/2

360 degs. = (1,0) 2pi


sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y

SOHCAHTOA:

S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)

C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)

T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)

*the hypotenuse is opposite the right angle.

*A= 1/2 bh*

*To find the area of a non right triangle use this formula:

*A= 1/2 (leg)(leg)SIN(angle b/w)

*When you have a non right triangle that has pairs, use the law of sines:

Sin A/a = Sin B/b= Sin C/c

*All you are doing is setting up a proportion.

**Remember to solve for an angle, you have to take the inverse.

*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)

*use an angle to orient yourself like SOHCAHTOA!!

Hope yall had a good thanksgiving holidays!!

Taylor's Reflection (( 22 November))

i missed this blog. OOPS! i totally forgot. i guess i was looking forward too much to the holiday. but ill make up the missed blog now.

i think i understand all there is that goes along with the graphing we learned last week
if i just make sure i remember and follow the steps one by one carefully it really isnt a hard concept to grasp.

i do have a little blip of confusion when it comes to distinguishing wether there is a horizontal shift or a vertical shift.

i think i just need an explanation of what to look for to decipher which shift is necessary

any help though would be wonderful!
thanks

Taylor's thanksgiving week reflection

So basically i remember everything we have learned so far in the latest weeks

Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)

Area of Right Triangles
A=1/2bh

SOHCAHTOA
sinΘ=opposite leg/hypotenuse
cosΘ=adjacent leg/hypotenuse
tanΘ=opposite leg/adjacent leg

Law of Sines
sinA/a - sinB/b = sinC/c
(used when you know pairs or opposites in a non-right triangle)

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°

Area of Inscribed Shapes
A=nr²sinΘcosΘ

All im worried about is completing the nine study guides in time for the exams although they will help me get a super good grade.
i predict these upcoming weeks to be the three most killer week of the school year.
everything is due
then its exams
my motivation will be the holidays just three weeks away

as soon as i get the study guides ill be sure to post all my questions

Stephen's Reflection from last week

Ok so i didnt think about school all this week so i forgot about the whole blog and comment thing so ima come do this right now to make it up. Im going to talk about something that gives me problems and is from chapter 1....graphing parabolas. Its not so much that its hard it just that i forget how to find x intercept, y intercept, vertex and axis of symmetry. so ill talk about this as best as i can.

There are 7 steps to this: 1. finding what way it opens, 2. how many intercepts, 3. finding the x intercepts, 4. finding y intercepts, 5. finding the axis of symmetry, 6. finding the vertex, and 7. graphing the parabola. to find the opening, you have to look infront of the x^2...if the number is positive then it opens up, if negative it opens down.
To find how many intercepts you plug in the formula b^2-4ac.
I kinda dont understand how to find x intercepts but i think you do completing the square .
when finding the y intercepts, you plug 0 in for x and solve for y.
to find the axis of symmetry you plug in the formula x=-b/2a
for the vertex you plug in your answer of the axis in for x and solve
then you just plug your answers in on a graph and you should get a parabola, if you dont know how to do this then you can use ur calculator

As said before im a little fuzzy on how to do these steps so if someone would be able to help or correct what i wrote if wrong then i would be grateful.

Stephen's Reflection

Ok so guess what...school tomorrow :( but oh well. I have been swamped with basically soccer with the practices and games and the tourny so I barely remember anything and im not ready to go back to school and learn more. But i do remember law of sines so i guess i will talk about that.

Law of sines works for non right triangles and it only works if you have a angle/side pair. The formula for law of sines is sinA/a=sinB/b=sinC/c. Big letters are the angles and little letters are the sides in triangle ABC. If you have angle A, angle B and side a then you will look for the side opposite of angle B. you will cross multiply to where ur problem will look like: bsinA=asinB. Then you divide by sin A and solve completely using ur calculator. Then you found your side b.

For what i need help with is remembering law of cosines so that way that bridge thing will go easier and im looking at my notes and im lost for some reason so i would be thankful if someone helped me.

Stephanie's Reflection

Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)
EG: The area of Triangle PQR is 15. If p is 5 and q is 10, find all possible measures of R.
A=15
P=5
Q=10
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R≈36.870°,143.130°

Right Triangles

• A=1/2bh
• SOHCAHTOA
• sinΘ=opposite leg/hypotenuse
• cosΘ=adjacent leg/hypotenuse
• tanΘ=opposite leg/adjacent leg

Law of Sines
sinA/a - sinB/b = sinC/c

• used when you know pairs in a non-right triangle
• you are setting up proportions

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443

Area of Inscribed Shapes
A=nr²sinΘcosΘ

Reflection #....um....Terrio's Reflection

Yeah so this week sucked! I did not have any time to do anything fun. The highlight of my week was going to Alicia's huge house and spending two full days to work on our bridge, but we finished it =]. The worst part of the week was the soccer tournament. Not only did I get in a fight with two guys on one team, but I could not walk after the first two games...and we played four. Well back to math.

So right now the only thing I really remember is the area of a non right triangle.

Formula:1/2(leg)(leg)sin(angle b/w)

So if you had 1/2(3)(6)sin(52) your answer would be: 7.100

Now for what I don't remember. I could really use some help remembering is the Law of Sines, and the Law of Cosines. Happy Thanksgiving...well late Thanksgiving =]

Reflection 15

Even though we did not have school this week you are to reflect back on an older concept. This is in preparation for your mid-term. As we talked about in class, pick anything from the year and write your reflection based on that.

Amy's Reflection #15

well obviously, we didn't learn anything new this week...so here's are some of the trig functions we learned so far...

pythagorean theorem: c^2 = a^2 + b^2 (use to find the hypotenuse of a right triangle)

area of a right triangle = 1/2 bh

SOHCAHTOA (used to solve right triangles):

Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg

area of a non right triangle = 1/2 (leg)(leg)SIN(angle b/w)

Example: non-right triangle: HIJ (left to right)H = 65 degrees, j = 2, i = 6. Find the area.

A = (1/2)(2)(6)sin(65)

A = 5.438

area of an isoceles triangle = bh

Law of Sines(used to non-right triangles):

Sin A/a = Sin B/b= Sin C/c

Example: you have a triangle with the sides 4 and 5 & you also have an angle of 30 degrees.

A = 1/2 (4) (5) Sin 30 degrees

A = 10 Sin 30 degrees which is aproximately = 5

Law of Cosines (used when you can't use Law of Sines):

(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)

Example: you have a triangle with the sides of 5, 6, and 7. find the angle between 5 and 6.

7^2=6^2+5^2-2(5)(6)

cos a7^2-6^2-5^2= 2(5)(6)

cos acos a= 7^2-6^2-5^2 / -2(6)(5)

a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))

a= 78.463 degrees

what i still don't understand is how you solve a whole trigangle using law of cosines..if someone could help me with that that would be great...!!!

Amy's Comments

ok since nobody asked an questions im gonna just give two examples...

Example #1: Find the angle of inclination...

2x+5y=15

m=-2/5

tan=-2/5

alpha=tan^-1(-2/5)

Example #2: For a triangle with C=36 degrees a=5 b=6, find side c....

c^2=5^2+6^2-2(5)(6)

cos36c=sqrt(25+36-2(5)(6)cos36)

c= 3.53

i hope this might help someone...& Happy Thanksgiving Everyone!!!

Alicia's Reflection #14

Okay, this week we started chapter 8 Trigonometry equations. Here are some formulas to remember:

Solving for a line:

m=tan alpha

Solving for a conic:

tan 2alpha= B/A-C

If A=C then alpha = pi/4 (always)

m=slope

alpha=angle of inclination

Example:

Find the angle of inclination of
x^2-2xy+3y^2=1

tan 2alpha=B/A-C

A=1 B=-2 C=3 (tan is +ve in 1&3)

tan 2alpha= -2/1-3=1

tan 2alpha=1

2alpha=tan^-1(1)

2alpha=45,225 degrees

2alpha=45/2, 225/2

alpha= pi/4


Happy Thanksgiving!

Amy's Rlection #14

ok this week we started chapter 8. Here's an explanation for 8.1:

1. For any line : m = tan (alpha)
**m = slope , (alpha) = angle of inclination

2. For a conic: tan 2 (alpha) = B/A-C

**if A=C then pie/4 (always)
**A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2

Examples:

1. Find the angle of inclination of x^2 - 2xy + 3y^2 = 1.

tan 2 (alpha) = B/A-C

A = 1 , B = -2 , C = 3

tan 2 (alpha) = -2/1 -3 = 1

tan 2 (alpha) = 1

2A = tan^-1 (1)

**if the # you're taking the inverse of is positive then find where it's negative and of course if it's negative then find where it's positive

**tan is negative in quad. #1 & 3

2 (alpha) = 45 , 225

alpha = 45/2 , 225/2

alpha = 22.5 , 112.5

2. x^2 + y^2 - 3xy + 4x - sqrt. x = 1

alpha = 1 (because A = 1 & C = 1 so A = C)

Helpful Hints:

quad #1: solve inverse
quad. #2 : make negative and add 180
quad. #3 : just add 180
quad. #4 : make negative and add 360

can someone tell me how you find the x-points and the phase shift. i know you need add something but what..i dont know..someone please help :(

Stephanie's Reflection

SOLVING TRIG EQUATIONS

for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2

Eg: find the value of x between 0 and 2pi for which sinx=6.
xsin-1=6
x=36.870, 143.130

2cos theta+9=7
theta=cos-17-9//3
theta=48.190
theta=131.810, 228.190

2tan theta+1=0
theta=tan -1 -1/2
theta=26.465
theta=153.435
153.435 times pi/180
theta=.853 pi

Stephen's Reflection

Ok so this week was basically a review for the test on Tuesday. I think that everyone basically has it down pack with how to do SOHCAHTOA, law of sines, and law of cosines. But what might confuse some people is when SOHCAHTOA, law of sines, and law of cosines is actually used. To start, SOHCAHTOA is used when you have a right triangle. This is the only formula that you use on a right triangle. Then for law of sines, there must be a pair. By a pair i mean that an angle has to have an opposite side and visa versa. If there are two of those then law of sines is used. And last for law of cosines. Law of cosines is used when you have a two legs and an angle inbetween. For law of cosines you would usually try to find all angles and sides.

There is one thing im a little confused about. I get a little confused when it askes me to find the are of a parallelogram. I know that you have to cut the parallelogram into two triangles and find the triangle's angles and sides if not listed but what im confused about is when you actually cut an angle because then you cant just cut it in half and so i get confused about how to find that angle of that specific triangle.

Taylor #13

This week we mostly reviewed the stuff we had previously learned in the chapter. We also learned about the law of cosines. I feel like I understand it. I think the only trouble I am having with it is sometimes when I type it into the calculator I get an error over and over again.
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!

Taylor #13

This week we mostly reviewed the stuff we had previously learned in the chapter. We also learned about the law of cosines. I feel like I understand it. I think the only trouble I am having with it is sometimes when I type it into the calculator I get an error over and over again.
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!

Taylor #13

This week we mostly reviewed the stuff we had previously learned in the chapter. We also learned about the law of cosines. I feel like I understand it. I think the only trouble I am having with it is sometimes when I type it into the calculator I get an error over and over again.
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!

Devin's Reflection #

This week we took another deadly step further into trigonometry. We learned more ways to solve for angle of a triangle, and also how to solve for the sides of a triangle. We also learned some formulas and ways to solve for the area of a triangle. To find the area of a right triangle,

1. Be sure that the triangle is right
2. Solve for the sides of the triangle
3. Use the formula : bh/2

To find the measures of a right triangle, use SOH CAH TOA

Sine= opp side/hyp

Cosine= adj side/hyp

Tangent= opp side/adj side

To find the measures of non-right triangles, you can use multiple techniques.
If the triangle has a pair (the measure of an angle and corresponding side), use The Law of Sine

sinA/a=sinB/b=sinC/c

If the triangles has no pairs, then use
The Law of Cosine

c^2=a^2+b^2-2(a)(b)cos(theta)

To find the area of a non-right triangle use the formula:
1/2(leg)(leg)sin(theta)

Alicia's Reflection #13

heyy, so this week was pretty much a review for our chapter 9 test on tuesday. We had a take home quiz for chapter 9 that was pretty easy once we learned the law of cosines. I do really good with those kind of questions but the minute i see a word problem its like i completely forget how to solve them. i need to figure out a way to know what method to use based upon the information that the word problem gives. if anyone has any tricks to know how to solve the word problems, it would help me out alot for the test on tuesday. Also, if anyone can help me with # 8 and #10 on the chapter test it would help me out greatly.

Review of chapter 9 Right triangles:

*the hypotenuse is opposite the right angle.

*A= 1/2 bh

*SOHCAHTOA---stands for....

Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg

*To find the area of a non right triangle use this formula:

*A= 1/2 (leg)(leg)SIN(angle b/w)

*When you have a non right triangle that has pairs, use the law of sines:
Sin A/a = Sin B/b= Sin C/c

*All you are doing is setting up a proportion.

**Remember to solve for an angle, you have to take the inverse.

*To solve a triangle with no angles, use the Law of Cosines:

(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)

*use an angle to orient yourself like SOHCAHTOA!!

Goodluck to everyone on the test!!

Stephanie's Reflection

Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)
EG: The area of Triangle PQR is 15. If p is 5 and q is 10, find all possible measures of R.
A=15
P=5
Q=10
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R≈36.870°,143.130°

Right Triangles

• A=1/2bh
• SOHCAHTOA
• sinΘ=opposite leg/hypotenuse
• cosΘ=adjacent leg/hypotenuse
• tanΘ=opposite leg/adjacent leg

Law of Sines
sinA/a - sinB/b = sinC/c

• used when you know pairs in a non-right triangle
• you are setting up proportions

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443

Area of Inscribed Shapes
A=nr²sinΘcosΘ

I don't understand how you know when to use Law of Sines or Law of Cosines. I especially don't understand how to plug them in (for some reason). Also, getting the Area of Inscribed Shapes eludes me.

Amy's Reflection #13

ok this week went by pretty quick. i guess it's because we had monday off. well anyway, we didn't really learn much this quick. the new formula we did learn this week was the Law of Cosines.

Law of Cosines:

(opp. side) = (adj. side)^2 + (other adj. side)^2 - 2(adj. side)(other adj. side) cos (angle between)

*you use the law of cosines when you can't use law of sines
* in other words, when you have: a non-right triangle, an angle measure with an unknown opposite side, and two side lengths on each side of that angle measure or when you have all three sides and no angle

Example1 (when given an angle & looking for the 3rd side):

you have a triangle ABC, angle C is 32 degrees, a = 5, and b = 6. Find c.

first you plug the values into the formula..

x^2 = 5^2 + 6^2 - 2(5)(6) cos (32 degrees)

x = sqrt of ((6^2 + 5^2 - 2(5)(6) cos 32 degrees))

x = 3.181

Example2 (when given the sides & looking for the angle in b/w):

you have a triangle with the sides of 5, 6, and 7. Find the angle in between 5 and 6.

7^2=6^2+5^2-2(5)(6)cos a

7^2-6^2-5^2= 2(5)(6)cos a

cos a= 7^2-6^2-5^2/-2(6)(5)

a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))

a= 78.463 degrees

*when you plug it in your calc. make sure you put two (( in the beginning and two )) in the end because if you don't you won't get the right answer...

ok now for what i dont understand....you know problems we get this weird shape and we have to make two triangles out of it..what do you after that? i also don't understand some of these word problem we get for homework like #s 17, 18, &19 on page 355..if anyone could help me with these problems that would be awesome..

AHHHHHH! I'm falling so far behind in this class now. As soon as we learned the Law of Cosines and I saw how much work it is I quit feeling like doing homework and I didn't feel like doing that whole quiz. I'm too lazy for this class. At least I know how to do the Law of Cosines I guess.

Law of Cosines:

(opposite side)^2=(adjacent leg)^2 + (other adjacent leg)^2 -2(adj leg)(other adj leg)cos(angle between the legs)

So all you really do is plug in the numbers from the triangle and then solve for whatever you need to solve for haha. When you plug in the numbers you just square root the whole thing then solve however you're suppose to be solving it.

OK. I need help with the Law of Sines. I have a feeling I was sleeping in class or something when we learned this because I do not remember a single thing about it and I can't understand my notes either.

Amy's Comments

1.) C = 90 degrees, b = 7, c = 12

angle A
cos A = 7/12
A = cos^-1(7/12)
= 54.315 degrees

angle B
sin B = 7/12
B = sin^-1(7/12)
= 35.685 degrees

length of a
tan 54.315 = a/7
(7) tan 54.315 = a/7 (7)
a = 7 tan 54.315
a = 9.747

area
A = 1/2 bh= 1/2 (7) (9.747)= 34.115

2.) two sides of a triangle have the lengths 7cm & 4cm. the angle between the sides measures 73 degrees. Find the area of the triangle.

1/2 (7) (4) sin (73) = 13.388 cm^2

Stephen's Reflection

Ok so this week we are in Chapter 9 and i think its pretty easy but at times i think i understand something but then i get confused. But anyway we did stuff SOHCAHTOA, law of sines, and area of non right triangles. I think that the easiest thing for me was the law of sines because i understand that completely.

The law of sines is:
sinA/a=sinB/b=sinC/c
This is for any triangle and it basically says angle A over side a=angle B over side b=angle C over side c. If you have a triangle ABC and it gives u the angle for A and B and gives you side b only then you would want to find side a which is directly across from angle a. so you do sinA divided by side a = sinB divided by side b and then you cross multiply and divide and your answer will be a=bsinA/sinB.

I understand basically all everything. Theres not much i dont understand but i do get a little confused on what formula to use out of all that we were given like on the ch. 9 test stuff but other than that im good.

TERRIO REFLECTION #12

OK so I did have a little more trouble this week than last. The beginning of the week was pretty easy but as we went on it got harder. On thing I really picked up was SOHCAHTOA. This makes it so much easier to remember the sin, cos, tan formulas. So here's what SOHCAHTOA means.

SOHCAHTOA:
SOH: sin=opposite leg/hypotenuse
CAH: cos=adjacent leg/hypotenuse
TOA: tan=opposite leg/adjacent leg

You use this stuff for right triangles. You use this to either find one leg or the hypotenuse, or you can use it to find angle measures. If you have the hypotenuse and one angle measure other than the right angle you would use either sin or cos to find one of the other legs, depending on which leg you want to find. If you have both the legs and the hypotenuse you can use SOHCAHTOA to find the angle measures too.

I did have a good bit of trouble finding the area of a non right triangle.

Alicia's Reflection #12

Okay so this week we had a quiz on 9-1 which was all word problems. I did not do well on it at all. I guess word problems are just not my thing. I could use some help with problems like on our homework worksheet (ch. 9 quiz). I do not understand how to set up the ones that do not give you any angles. If anyone can help me with this it would help me out tremendously. Lets review some trigonometry.

--The hypotenuse is opposite the right angle.

--A= 1/2 bh

SOHCAHTOA---stands for....

Sin θ= opposite leg/hypotenuse

Cos θ= adjacent leg/hypotenuse

Tan θ= opposite leg/adjacent leg

--To find the area of a non right triangle use this formula:

--A= 1/2 (leg)(leg)SIN(angle b/w)

--When you have a non right triangle that has pairs, use the law of sines.

Sin A/a = Sin B/b= Sin C/c

All you are doing is setting up a proportion.

**Remember to solve for an angle, you have to take the inverse.

Trigonometry Pt. 2

If you are given a right triangle like shown above, you can find the missing values (side a and angle B) with a little bit of simple trigonometry and geometry.

Using SOH CAH TOA you should be able infer that sin(30.7) = (a/c)

Since c is 10.5, 10.5sin(30.7) = a

So a≈5.3607

You've solved for the sides

Now, subtract the two angles from 180° to find your final angle.

B=59.3°

Taylor Reflection #12

Everything we learned this week was very simple formulas to memorize and plugging in to solve.
I also took the chapter seven test finally
I NEED SOMEONE TO SHOW ME HOW TO SOLVE WHAT WE WILL NEED TO SOLVE WITH OUT THE CALCULATOR if that makes sense,
i dont know if that makes sense but in other words... i need someone to show me how to work the problems that i work now with a calculator that i wont have a calculator for later because everthing from chapter seven i use a calculator for..

I ALSO COULD USE HELP ON APPLYING THE UNIT CIRCLE.
ive learned it and i kind of know how to apply it
but i don't know all of the times when it is applied..


What i do know is how to clear up any possible confusion for plugging into the formulas of this past week ::
to solve a right triangle you use SOHCAHTOA

The hypotenuse is opposite of the right angle.
The adjacent leg is the side that is attached to the "given angle (not right angle)"
The opposite leg is the side that is across (or opposite) the "given angle (not right angle)"

SOHCAHTOA is really a anagram to remember the way to work out a right angle triangle problem using sine cosine and tangent

SOH: sin(theta)=opposite leg/hypotenuse
CAH: cos(theta)=adjacent leg/hypotenuse
TOA: tan(theta)=opposite leg/adjacent leg

solving for a right triangle is just a matter of drawing and plugging in





#2 Laws of Sines: sin A/a = sin B/b = sin C/c

*used when pairs are known in non-right angles
*set up a proportion

Must use a "pair" for sin A over a

a "pair" is given after you take the given information from a problem and you look for an angle given and a number to represent the leg directly across from that angle. Sin B/b is solved by simple algerbra and Sin C/ c is what you"ll get to have the measurements for the rest of the triangle.



#3 Area of a non-right triangle = 1/2 (leg) (leg) sin (angle b/w)

This can possibly be the most simple formula to remember and use
legs are numbers for either the hypotenuse, the adjacent, or the opposite lines in the drawn problem. any "leg" will work for the problem.

Amy's Reflection #12!

here are some things i understand:

to solve a right triangle you use SOHCAHTOA

Area of a right angle = 1/2 bh

Laws of Sines: sin A/a = sin B/b = sin C/c
*used when pairs are known in non-right angles
*set up a proportion

Area of a non-right triangle = 1/2 (leg) (leg) sin (angle b/w)

Examples:

1. B = 90 degrees C = 50 degrees b = 10

Angle A
180 - 90 - 50 = 40

Side a
cos 50 = a/10
a = 10 cos 50
a = 6.428

Side c

sin 50 = c/10
c = 10 sin 50
c = 7.660

area = 1/2 (6.428) (7.660) = 24.620

2. two sides of a triangle have the lengths 7cm & 4cm. the angle between the sides measures 73 degrees. Find the area of the triangle.

1/2 (7) (4) sin (73) = 13.388 cm^2

Hints (word problems):
if it mentions anything about anything that has height or something on the ground that means you're gonna draw a right angle and solve...and the other problems you're gonna draw an isosceles triangle and remember to draw down the middle to make it two separate triangles...
(i hope that makes sense....)

the worksheets are due tuesday, huh?

Stephanie's Reflection

AREA OF A NON-RIGHT TRIANGLE
A=1/2(leg)(leg)sin(angle between)

EX: 2 sides of a triangle have legs 7cm and 4cm. the angle between the sides measures 73 degrees. find the area of the triangle
1/2(7)(4)sin(73)=13.388com^2

the area of triangle PQR is 15. if p is 5 and q is 10, find all possible measures of R
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin^-1(3/5)
R=36.870, 143.130

RIGHT TRIANGLES

• hypotenuse opposite angle
• a=1/2bh
• SOHCAHTOA

sin theta=opposite/hypotenuse
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent

LAW OF SINES
sinA/a sinB/b sinC/c

• used when you know pairs in a non-right triangle
• you are setting up proportions

EG:
sin115/123X=sin theta/16
123sin theta=12sin115
sin theta=16sin115/123
theta=sin^-1 16sin115/123
=6.771

sin30/4=sin15/x
xsin30=4sin15
x=4sin15/sin30
x=4sin15/.5
x=2.071

A civil engineer wants to determine the distances from points A and B to an inaccessible point C. From direct measurements, AB=25cm, CA=110 and angle B=20. Find AC and BC.
sin50/25=sin20/b
bsin50=25sin20
b=25sin20/sin50
b=11.162m
sin50/25=sin110/a
asin50=25sin110
a=25sin110/sin50
a=30.667

Stephen's Reflection

Ok so this week we finished up with trig and did the test on thursday. Hopefully everyone did well on it and i also found that it was kinda easy. Maybe because i actually studied for this one. Anyway after the test we started Chapter 9 with SOHCAHTOA. This is fairly easy when you learn what the sides of a right triangle is.

The hypotenuse is opposite of the right angle.
The adjacent leg is the side that is opposite of the hypotenuse.
The opposite leg is the side that is opposite the top angle.

What SOHCAHTOA really means is:
SOH: sin(theta)=opposite leg/hypotenuse
CAH: cos(theta)=adjacent leg/hypotenuse
TOA: tan(theta)=opposite leg/adjacent leg

When given a right triangle, all you have to do is plug it in. If it only gives u a hypotenuse and an opposite leg, you would use sin(theta). If it gives u a hypotenuse and adjacent leg, you will use cos(theta) and if it only give you the opp leg and adj leg, you will use tan(theta).

That is basically it for SOHCAHTOA. Theres not much that i dont understand because i find trig pretty easy and this is fairly easy as well.

Taylor Reflection #11

This week was okay i guess
im still trying very hard to catch up on what i missed
i am facing the dreded chapter seven test this week
and im very nervous because although i got to review with the class i didnt get to experience any quizes of this chapter or hear and learn the information first hand
i think i might just be okay
i have learned and memorized all the formulas and rules
however im still trying to practice the application of the rules and formulas
and whats funny is i completely understand chapter nine
i guess it will all come down to doing all the problems the book has to offer and hoping for the best

what i definitely understand is the trig chart and i also made up a quick way to remember it
what i also know is the methods
such as what the variables for the four word problem formulas stand for

R- radius
((distance will mean radius))
S- arc length
Theta- angle
(( apparent size will mean Theta)
K- area of a sector

Take the word problems slowly
take the time necessary to discover what is given in the problem and work step by step from there
just tell your self there is a way to solve this problema and it cant be too hard



what i think i need help on will really be the key way for recognizing which problem will be worked with which method.
any tricks to help me decipher when i take the test would be awesome

Thanks!

Stephanie's Reflection

Unit Circle

sin theta = y/r

cos theta = x/r

tan theta = y/x

csc theta = r/x

sec theta = x/y

cot theta = x/y

r=sqrtx^2+y^2

90 pi/2
180 pi
270 3pi/2
360 2pi

sin pos in one and two and neg in three and four

cos pos in one and four and neg in two and three

tan pos in one and three and neg in two and four

cot pos in one and neg in two three and four

sec pos in one and neg in two three and four

csc pos in one and two and neg in three and four

sec = 1/cos

cos = 1/sin

EG: 3^2+3^2=5

sqrt25=h

r=5
y=3
x=4

sin theta=3/5

cos theta=4/5

tan theta=3/4

csc theta=5/3

sec theta=5/4

cot theta=4/3

find the reference angle for sin 210 degrees
sin 210-__-___sin_30___
210-180=30
sin 210=-1/2

find the reference angle for cos 315 degrees
cos 215=__+___cos__45___
cos 315=+/-sqrt2/2
315-180=45

tan 695=__-__tan__25__
tan 695=-466
695-360=335

c=2pir
2pir=1/2 2piR
r=1/2R
cos theta-r/R
=1/2
theta=60

Alicia's Reflection #11

Okay so this week we had our chapter 7 text on trigonometry. I hope everyone did good on the test. I thought I did alright except for the 3 word problems on latitude and earth and all that so I may need some help with those types of word problems. We also started chapter 9 this week on Right triangles.

Notes:

*the hypotenuse is opposite the right angle.

*A= 1/2 bh

*SOHCAHTOA---stands for....

Sin θ= opposite leg/hypotenuse

Cos θ= adjacent leg/hypotenuse

Tan θ= opposite leg/adjacent leg

Example: a right triangle has an angle of 28 degrees, an opposite leg of 40 degrees, and an adjacent leg b.

Tan θ= opposite leg/ adjacent leg

Tan 28 degrees=40/b

b Tan 28 degrees=4

b Tan

Yeah so I'm diggin' this trigonometry stuff. I think I really am picking up this stuff a lot better than the stuff we were learning the first quarter. One thing this week that I learned was Right Triangles. When I saw this I was surprised that I already knew how to do it. I learned it in Geometry last year. I even remembered SOHCAHTOA

Notes:
hypotenuse opposite right angle
A=1/2bh
SOHCAHTOA

A right triangle has an angle of 32 degrees. The adjacent leg is 5. Find the hypotenuse.

Answer=cos32=5/x
xcos32=5
x=5/cos5 or 5/5.896

The only thing that I'm still having trouble with is the word problems with the arc length and the planets and stuff.

Alaina's Reflection

This week was pretty easy, I think. This week, we learned to use trig functions to find the measurements of the sides and angles of an isosceles triangles.
What happened to chapter 8, by the way?

1. drop down from the top of the triangle to create the altitude and two right triangles.
2. divide the base in half to give you the base measurement of both right triangles.
3. to find you altitude, your newly found leg, you would use the Pythagorean Theorem.
4. to find the angle measures, you would use trig inverses. (sin^-1, cos^-1, and tan^-1)

As far as things that I don't understand, there isn't much. Actually, I can't think of anything at the moment that I don't understand that has to do with Trig. The test on Thursday was pretty easy, mainly because I studied my butt off and had it retaught to me about four times. So I understand pretty much all that I've learned so far concerning Trig.

An Introduction to Triangle Trigonometry

Okay, here we are, basic triangle trigonometry.  Here are a few things that will come in handy:

Trigonometric Properties:

SOH CAH TOA 

sine=((opposite)/(hypotenuse))

cosine=((adjacent)/(hypotenuse))

tangent=((opposite)/(adjacent))

And the less commonly used: CHO SHA CAO

cosecant=((hypotenuse)/(opposite))

secant=((hypotenuse)/(adjacent))

cotangent=((adjacent)/(opposite))

~ Use the functions above when solving for a side

~ Use the inverses of the functions above when solving for an angle.

Naming Parts of a Triangle:

A triangle’s sides are labeled as lowercase letters.

A triangle’s edges are labeled as either capital or greek letters (alpha, beta, gamma, delta, phi, psy, omega, theta, etc).

Geometric Properties:

The sum of the angles of a triangle add up to 180°. In other words, if you know 2 angles of a triangle you can subtract those from 180 to find the third angle!

The sum of the angles of any polygon with 4 or more sides add up to 360°.

Hope this helps!

Amy's Reflection #11

ok ths week started chapter 9..i wonder if we're gonna go back and do chapter 8??..anyway im not really good at solving right angles so im not..dont want solve a problem wrong and confuse someone...but here are some things do understand:

Coversion:

Degrees to Radians is: degrees*pi/180
Radians to Degree is: radians*180/pi

Unit Circle:

90 degs. = (0,1) pi/2
180 degs. = (-1,0) pi
270 degs. = (0,-1) 3pi/2
360 degs. = (1,0) 2pi

sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y

SOHCAHTOA:

S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)

C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)

T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)

and i still need with the inverse stuff..if someone could help me with that..i'll really appreciate it..