Chapter 10

Chapter 10 is strictly formulas. It may not be easy to memorize all of them, but once you do this will be an extremely easy chapter. You will also need the trig chart again in this chapter! Today I'm going to explain section 10-1's formulas.

They are:

1. cos(alpha+/-beta)=cos alpha cos beta-/+sin alpha sin beta

2. sin (alpha+/-beta)=sin alpha cos beta+/-cos alpha sin beta

3. sin x+sin y=2sinx+y/2 cos x-y/2

4. sin x-sin y=2cosx+y/c sin x-y/2

5. cos x+cos y=2cosx+y/2cosx-y/2

6. cos x-cos y=-2sinx+y/2 sin x-y/2

Now for an example.(REMEMBER: in this chapter you REPLACE, NOT PLUG IN!)

Find the exact value of Sin 15
1. Since sin 15 isn't on the trig chart, you must use a formula that will make it.
2. So you will use Sin(45-30)
3. You will now get, Sin45 cos 30-cos 45 sin30
4. Using the trig chart, you get Sqrt 2/2 x sqrt 3/2 - Sqrt 2/2 x sin 1/2
5. Simplify. You get (sqrt 6-sqrt 2)/4

That's how you do thattt :)

Carleyyy :)

10-3 Double-Angle and Half-Angle Formulas

For the following problem, one would use the formula Cos2alpha = cos^2alpha - sin^2alpha, 1-2sin^2alpha or 2cos^2alpha - 1

2 cos^2 pie/8 - 1
Notice that pie/8 is in the place of alpha, which means pie/8 is your alpha. In this chapter all you are doing is replacing so you would replace that part of the formula with what it is equal to; though first you would want to change pie/8 to degrees and get 45degrees. Once this is done you replace with the formula and get cos 2(45) and that is equal to cos(90), which is equal to 0. (Final Answer)

~ Parrish Masters

10-1

soooo I left my binder at school friday because I was too worried about getting out of that stupid, but educational, building that consist of dull colors and smelly children. Sooo im going to wing it. I have all the formulas on a notecard so i think that will help a little bit.

So here it goes, im going to show you all of the formulas then i might work a few problems.

1. cos(alpha+/-beta)=cosalphacosbeta-/+sinalphasinbeta

2. sin (alpha+/-beta)=sinalphacosbeta+/-cosalphasinbeta

3. sinx+siny=2sinx+y/2 cosx-y/2

4. sinx-siny=2cosx+y/c sin x-y/2

5. cosx+cosy=2cosx+y/2cosx-y/2

6.cosx-cosy=-2sinx+y/2 sinx-y/2

Soo now that i have showed you the formulas im going to give you 2 examples. Make sure you follow closely cause im only going to show you how to do it once!

Ex1. a) cos 120 cos 30 + sin 120 sin 30=
cos (120 - 30)=
cos (90)=
0

Ex2. a) sin70 cos20 + cos70 sin20
sin(70+20)=
sin(90)
1

love yours truely <3

10-1

10-1 has plenty of formulas.

Formulas:

1. Cos(alpha+-beta)=CosalphaCosbeta-+SinalphaSinbeta
2. Sin(alpha+-beta)=SinalphaCosbeta+-CosalphaSinbeta
3. Sinx+Siny=2Sin x+y/2Cos x-y/2
4. Sinx-Siny=2Cos x+y/2Sin x-y/2
5. Cosx+Cosy=2Cos x+y/2Cos x-y/2
6. Cosx-Cosy=-2Sin x+y/2Sin x-y/2

Example 1: Find the exact value of sin150 degrees.

• Step one: Look on the trig chart and see if any of the degrees add up to give you 150 degrees. 90+60=150
  
• Step two: Place 90 and 60 in the formula, sin(alpha+-beta)=sinalphacosbeta+-cosalphasinbeta
• Step three: Sin(90+60)=Sin90Cos60+Cos90Sin60

And after that just solve it and get the exact value.

Amber :)

so many formulas, so little time

Chapter 10 is formula after formula and the trig chart has come back to haun-I mean prove its usefullness. I am going to talk about the formulas from 10-1. Here we go.

1. cos(alpha+/-beta)=cosalphacosbeta-/+sinalphasinbeta

2. sin (alpha+/-beta)=sinalphacosbeta+/-cosalphasinbeta

3. sinx+siny=2sinx+y/2 cosx-y/2

4. sinx-siny=2cosx+y/c sin x-y/2

5. cosx+cosy=2cosx+y/2cosx-y/2

6.cosx-cosy=-2sinx+y/2 sinx-y/2

EXAMPLE :D

Find the exact value of sin75degrees.

sin(45degrees+30degrees)=sin45cos30-cos45sin30

45degrees is alpha and 20 degrees is beta. I simply plugged it in to equation 2.

That's the way you do that. Have a happy Halloween :D

10-3

This week I am going to teach you all how to work out trig problems by using formulas for sin 2a, cos 2a, tan 2a, sin a/2, cos a/2, and tan a/2. There are ten different formulas that we use for these types of problems. They are as follows:
• sin 2a= (2 sin a) (cos a)
• cos 2a= (cos^2 a) – (sin^2 a)
• cos 2a= 1- 2 sin^2 a
• cos 2a= 2 cos^2 a -1
• tan 2a= ( 2 tan a) / ( 1- tan^2 a)
• sin a/2= +/- sq. root of ((1 – cos a) / (2))
• cos a/2= +/- sq. root of (( 1+ cos a) / (2))
• tan a/2= +/- sq. root of (( 1- cos a) / ( 1+cos a))
• tan a/2= +/- (( sin a) / ( 1+ cos a))
• tan a/2= +/- (( 1- cos a) / (sin a))

Example:
Simplify cos^2 15 degrees – sin^2 15 degrees
• cos 2(15 degrees)
• cos 30 degrees
• (sq. root of 3) / (2)

-Braxton

10-1

10-1 Formulas for cos (a +/- B) and sin (a +/- B)

Sum and Difference Formulas for Cosine and Sine:
-cos (a +/- B) = cos a cos B -/+ sin a sin B
-sin (a +/- B) = sin a cos B +/- cos a sin B

Rewriting a Sum or Difference as a Product:
-sin x + sin y = 2 sin (x + y/2) cos (x - y/2)
-sin x - sin y = 2 cos (x + y/2) sin (x - y/2)
-cos x + cos y = 2 cos (x + y/2) cos (x - y/2)
-cos x - cos y = -2 sin (x + y/2) sin (x - y/2)

Example 1: Simplify the given expression.
a) cos 105 cos 15 + sin 105 sin 15
=cos (105 - 15)
=cos (90)
=0
b) sin 3x cos 2x - cos 3x sin 2x
=sin (3x - 2x)
=sin x

Example 2: Find the exact value of each expression.
a) cos 105
=cos (60 + 45)
=cos 60 cos 45 - sin 60 sin 45
=(1/2) (sq root of 2/2) - (sq root of 3/2) (sq root of 2/2)
=sq root of 2/4 - sq root of 6/4
=(sq root of 2 - sq root of 6)/2

-Jordan Duhon

10-1 Formulas for cos(a+-B) and sin(a+-B)

This week I am going to explain how to work problems using formulas for cos(a+ or - B) and sin(a+ or - B). There are 6 different formulas for these types of problems. These formulas are:

1. cos(a + or - B)=cosa cosB - or + sina sinB
2. sin(a + or - B)=sina cosB + or - cosa sinB
3. sinx+siny=2sin (x+y/2) cos (x-y/2)
4. sinx-siny=2cos (x+y/2) sin (x-y/2)
5. cosx+cosy=2cos (x+y/2) cos (x-y/2)
6. cosx-cosy=-2sin (x+y/2) sin (x-y/2)

Now I am going to work some examples using these formulas. There are many different ways to work these problems.

Example 1: cos 105 degrees

• You are going to use your trig chart to help work this problem.
• You are going to use formula 1 to solve this problem.
• Since 45 and 60 degrees are on the trig chart and they add up to equal 105, you are going to use those two degrees.
• cos(45+60)=cos45 cos60-sin45 sin60
• You then plug those into the trig chart.
• You answer is going to be: square root of 2 - square root of 6/4

Example 2: sin75 cos15 + cos75 sin15

• You are going to replace this with one of your formulas above.
• That formula above is the same as sin(a+B)
• Once you replace with that formula you are going to get sin(75+15)=90 degrees
• sin 90 degrees on the trig chart equal 1
• 1 is goin to be your answer

And that is how you work problems using formuals for cos(a+B) and sin(a+B)

--Halie! :)

10-1

today im showing you the formulas for cos (alpha +/- beta) and sin (alpha +/- beta). i will also show you an example on using them. the key word for this chapter is REPLACEING. this is because you are replacing things in the formulas. the first formula is cos (alpha +/- beta) = cos (alpha) cos (beta) -/+ sin (alpha) (beta). the next one is sin (alpha +/- beta) = sin (alpha) cos (beta) +/- cos (alpha) sin (beta).


EX: sin 75 cos 15 + cos 75 sin 15

sin(alpha + beta)= sin (alpha) cos (beta) + cos (alpha) sin (beta)

sin(75+15)= sin 90 = 1

Law of Sines

Law of Sines is used on non-right triangles. When you take the inverse, you will get two answers. You use Law of Sines when you now an angle and its opposite leg. You will have to cross multiply and divide in this section.

Formula: SinA/a = SinB/b = SinC/c

Example 1: Find the missing angle and the lengths of c and b.

A

C B
First, we have to find the missing angle. Since you already have two of the angles, all you have to do is subtract the two angles you already have form 180 to get the third angle.

180-40-80 = 60 degrees

Now, let's solve for c and b. This is when Law of Sines comes in.

Sin80/14 = Sin60/b

bSin80 = 14Sin60

b = 14Sin60/sin80

b = 12.311

Sin80/14 = Sin40/c

cSin80 = 14Sin40

c = 14Sin40/Sin80

c = 9.138

Example 2: Find angle A and the lengths of b and c.

A

C B

180-41-88 = 51 degrees

Sin41/20 = Sin88/b

bSin41 = 20Sin88

b = 20Sin88/Sin41

b = 30.466

Sin41/20 = Sin51/c

cSin41 = 20Sin51

c = 20Sin51/sin41

c = 23.691

9-4 law of cos

i am a little late on this blog but i rather get the 1/2 credit then the no credit so here we go. today im showing you the law of cosines. this is the last resort if law of sin does not work. the formula is opp leg^2=adj leg^2 + adj leg^2-2(adj)(adj) cos (angle b/w them). you can find angles or the other leg depending on what the directions are. usually it says solve and you will have to find everything so here is an example.


EX: solve the triangle

a=8 b=5 angle C=60 degrees


first you do the formula- c^2= 5^2 + 8^2 - 2(5)(8) cos 60
then square root everything and you get c=7
now find angle A- 8^2= 7^2 +5^2 -2(7)(5) cos A
move everything to the left and take in verse of cos A and u get A=81.787
now find angle B- 60+81.787= 141.787   then 180-141.787=38.213
now the triangle is solved

Sorry Folks

I would like to apologize sincerely from the bottom of my heart in the announcement that there will be no episode of "The Cory Teaches You How to Do Something Show" this week. MTV called and said they would like to negotiate a contract for the airing of the show on national television. In the contract it says i can not do a show for this week. So heres to a regular boring blog.

I dont remember learning anything this week and i dont have my binder yet again so im just going to do SOHCAHTOA

Formula:

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

Example: Using the picture above, say B= 38degrees a= 18 b= 26 Fnd A and c

First, lets find c. Do tan= opposite/ adjacent =26/18= 13/9
c=13/9

A= 180-90-38=52 degrees

Im sorry for this boring and dissapointing blog you guys but i have alot of crap to do for homework tonight and i have a test to study for tomorrow that i cant study for without my binder:) How Lovely!!!! GOodnight everybody

blog dismissed,bring in the dancin lobstahs.

So this was a four day week and we took a test on Thursday and question day was Wednesday and we had a quiz Tuesday and class work Monday so I do not recall learning anything new. That being said, I’ve decided that this week my blog will be on Law of Sines. It’s a rather simple concept. You use this for non-right triangles. This should be your first choice in solving the triangle because it is shorter and much less complicated than Law of Cosines.

Formula:

SinA/A=sinB/B=sinC/C

Solve triangle ABC if BC=3, A=62 degrees, and B=43 degrees

-First you draw your triangle. (As always you should write your answers in as you get them.)

To find C, simply subtract 180-62-43.

C=75 degrees

Next we’ll solve for AC.

Sin62degrees/3=sin3degrees/B

You’re gonna cross multiply. So Bsin62degrees=3sin3degrees

Solve for b.

B=3sin3degrees/sin62degrees

B=.178

Use the same process to solve for AB.

Sin62degrees/3=sin75degrees/c

Csin62degrees=3sin75degrees

C=3sin75degrees/sin62degrees

C=3.381

Your final triangle should look like this J

That’s that.

--Sarah

9-3 Law of Sines

This week I am going to teach you all how to use the Law of Sines. You can only use this on non-right triangles. You try to use the Law of Sines before you try to use the Law of Cosines. If you know an angle and an opposite leg, then you can use the Law of Sines.
When you take the inverse for one of these problems, you always will get two answers.
• After you plug the problem into your calculator and get the first angle, you multiply the angle by -1 and add 180 to get the second angle.
The formula for Law of Sines is: sin A/a=sin B/b=sin C/c
To solve, you have to cross multiply and divide.
Example: Solve triangle ABC if, BC 4, A=45 degrees, B=60 degrees, C=75 degrees
• First you have to draw the triangle.
• Next you have to set up an equation to solve for the first side:
sin 45 degrees/14=sin 75 degrees/c
• When you cross multiply and divide, it gives you: c=19.124
• Then you set up the next equation:
sin 45 degrees/14=sin 60 degrees/b
• That gives you b=17.146

-Braxton-

Law of Sines Part 2 (If there is a Part 1 I Don't Remember but I'm Sure I Made One B4)

Ok so, I'm going over law of sines again. Law of sines can only be used on non-right triangles. The formulas you use for this are: SinA/a = SinB/b = SinC/c

Say you have a Triangle ABC, which A=60degrees, a=4, b=3; you are told to solve the triangle. In a case like this, when you have two legs and at least one angle (of course, the angle has to be relevant to one of the legs), you would use law of sines to solve the triangle.

First thing you would do is set up the formula, so Sin60/4 = SinB/3, cross multiply using your calculator (3Sin60/Sin4); get B = 37.245
Now add B and A and subtract that number from 180 to find angle C and get 82.755.
Use law of sines to find c. Sine82.755/c = Sine60/4; get 4.582, so you end up with:
A= 60, B = 37.245, C = 82.755, a = 4, b = 3; c = 4.582

~ Parrish Masters

Law of Sinessssssss(:

SO guysss, we're gonna learn how to solve triangles using Law of Sines today.

(I'm also in a colorful mood, so be readddddy!)

You use Law of Sines on non-right triangles ONLY.

You will always get two answers when take the inverse.

You use this when you know an angle and an opposite leg!

The formula that you will use is SinA/a = SinB/b = SinC/c

You will cross multiply and divide things to solve.

Example timeeeee!

Find angle B!

Sin 25/2 = sinB/3

(by cross multiplying)

3sin25=2sinB

SinB=3Sin25/2

B=Sin^-1 3Sin25/2

B=39.341

Because sin is positive in quadrant I and II, make it negative and add 180.

So you get 140.659.

Your answers are B= 39.341 and 140.659.

I'll even go a step further and tell you how to find out how many triangles are possible...

You do this by adding each of the two angles you found with the one given in the picture.

39.341+25=64.341

140.659+25=165.659

Since they are both under 180º, their is two possible triangles :)

---------------CARLEY :)

9-3 Law of Sines

Law of Sines is only used when the triangle is not a right triangle.
You use Law of Sines when the triangle has an angle and it's opposite leg.
When you use the inverse, you will get two answers.

Formula: SinA/a = SinB/b = SinC/c

Ex. 1 Triangle ABC gives you angle A, angle B, angle C, and leg a. Angle A = 80, angle B = 25, angle C = 75, and leg a = 10. Solve the triangle.

• First, you need to see which angle has an opposite side: angle A and leg a are opposite sides. You will use this angle and leg to find the other legs.
  
• Second, you need to pick one of the legs to find (doesn't matter which one). To find leg b: Sin80/10=Sin25/b. bSin80=10sin25. b=10Sin25/Sin80. b = 4.291
• Now find leg c: Sin80/10=Sin75/c. cSin80=10Sin75. c=10Sin75/Sin80. c = 9.808

Ex.2 Triangle ABC gives you angle B, leg a, and leg b. Angle B = 28, leg a = 7, and leg b = 3. Solve the triangle.

• First, find out if angle B has an opposite side: It does. Leg b is its opposite side.
• Second, since you do not have any other angle besides angle B, you have to find the other angles first: Sin28/3=SinA/7 (use angle A because it gave you leg a which is opposite from angle A). 3SinA=7Sin28. (take the inverse of sin) B=sin^-1(7Sin28/3). B = .019
• Now find angle C: 180-28-.019 = 151.981
• After you find angle C, you can now find leg c: Sin28/3=Sin151.981/c. cSin28=3Sin151.981. c=3Sin151.981/Sin28. c = 3.002
  

9-3

9-3 Law of Sines

We use Law of Sines on non-right triangles. You can use this formula when the triangle gives you an angle an opposite leg to solve the triangle. You will get two answers when you find the inverse. You cross multiply using the following formula:
sin A/a = sin B/b = sin C/c

Example 1: Triangle ABC gives you Angle A = 60, Angle B = 95, Angle C = 25, and a = 8. Solve the triangle.

• sin 60/8 = sin 25/c


• c sin 60 = 8 sin 25


• c = 8 sin 25/sin 60


• c = 3.904


• sin 95/b = sin 60/8


• b sin 60 = 8 sin 95


• b = 8 sin 95/sin 60


• b = 9.202


• 180 - 90 - 60 = 25


• Angle C = 25

Example 2: Triangle ABC gives you Angle A = 76, a = 12, and b = 4. Solve the triangle.

• sin 76/12 = sin B/4


• 12 sin B = 4 sin 76


• B = sin^-1 (4 sin 76/12)


• Angle B = 18.871


• 180 - 18.871 - 76 = 85.129


• Angle C = 85.129


• sin 76/12 = sin 85.129/c


• c sin 76 = 12 sin 85.129


• c = 12 sin 85.129/sin 76


• c = 12.323

9-4 Law of Cosines.

So, this week I am going to explain how to solve triangles using Law of Cosines. This is done with one easy formula. This formula is:

• leg^2=adj. leg^2 + other adj. leg^2 - 2(adj. leg)(other adj. leg) cos(angle between the two adj. legs)

--Now that you know the formula you should be able to solve for these types of trianlges. So I am going to do a few examples.

Example 1: Solve for anlges A and B.

To solve for angle B, 7 and 8 are going to be your adjacent legs. Now your going to plug everything into the formula.

• 5^2=7^2 + 8^2 - 2(7)(8)cos(B)
• B=cos^-1(5^2-7^2-8^2/-2(7)(8))
• B=38.213 degrees

Now to solve for angle A, all you have to do is add up angle B and C and subtract from 180.

• A=81.787 degrees

Example 2: Solve for angles D and E and for side length f.

First lets solve for the side length f. This is very simple. Angle F will be your angle in between of the adjacent legs 9 and 5.

• f^2=9^2 + 5^2 - 2(9)(5)cos(115)
• Take the square root of both sides.
• f=12.001

Now we are going to solve for anlge D. This is going to be just like example 1.

• D=cos^-1(5^2-9^2-12.001^2/-2(9)(12.001))
• D=22.186 degrees

Add up angles D and F and subtract from 180 to find out what angle E is.

• E=42.814 degrees

And that is how you solve problems using Law of Cosines!

--Halie :D

9-4 Law of Sines

Law of Sines is be only used when solving non-right triangles. There will be but 2 answers for the inverse. It is can be used when u know an angle and an opposite.

The formula is

sinA/a=sinB/b=sinC/c

To solve all that needs to be done is cross-multiplication

Ex 1:

Triangle ABC angle A= 110 degrees angle C= 50 degrees Side c=25m

Solve the triangle:

Interior angles add up to 180 degrees in a triangle so add up 110 and 50 then subtract the sum from 180 to find angle B.

Angle B= 20 degrees

To find side a apply law of Sines.

sin 50 degrees (C)/25 m=sin 110 (A)/ side (a)

Cross multiply and you'll get 25 sin110=sin 50

So plug in 25 sin 110 and you'll get 23.492

So set that equal to sin 50 degrees

It should look like 23.492=sin 50

Divide sin 50 degrees on both sides

side (a)= 14.837m

To find side b apply law of Sines

b= sin 50 degrees(C)/25 m=sin 20/side(b)

Cross multiply again and you'll get 25 sin 20=sin 50 degrees

Divide both sides by sin 50 degrees

The answer comes out to be 11.162 m=side b

-Sameer

Yaaaay,cosines :D

Alright kiddie winks, today I'm going to teach you all about our friend the Law of Cosines :D

First off, you should know that this is sort of a last resort. Only use it when you can't do SOHCAHTOA or Law of Sines. The formula is somewhat lengthy, but make sure you know it because we have a quiz Tuesday.

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

Now, who's ready for an example?!

Solve.

So for starters, we'll use our equation to find the third leg.

x^2=8^2+5^2-2(8)(5)cos60degrees

Square root both sides and plug the equation into your calculator like this:

square root of (8^2+5^2-2(8)(5)cos60degrees

Round to three decimal places.

You should get 7. Go back to your graph and fill this in on the remaining side.

Next, we use our equation to find another angle. We'll solve for angle A.

8^2=7^2+5^2-2(5)(7)cosA

Move all squares to one side then divide by -2(5)(7)

8^2-7^2-5^2/-2(5)(7)=cosA

Take the inverse and plug the following into your calculator:

cos-1((8^2-7^2-5^2)/(-2(7)(5))

Keep in mind that I am not using the ( ) for my health, you actually need them when plugging into your calculator.

Your answer should be 81.787 degrees. Fill that into your triangle.

To find the angle B, just subtract A and C from 180 degrees.

180-81.787-60= 38.213 degrees.

And that's that.

--Sarahhhhhhhhhhhhhhhhhhhhh

9-4 Law of Cosines

This week I am going to teach you how to use law of cosines. You use it to solve non-right triangles. It is only used when Pythagorean and law of sines do not work. It is pretty easy. (Capital letters represent angles, and lowercase letters represent legs).
There is only one formula that you need to remember for this section:
(opposite leg^2) = (adjacent leg^2) + (other adjacent leg^2) – 2 (adjacent leg) (other adjacent leg) cos (angle between)
Example 1:
In triangle ABC, C= 60 degrees, b= 5, and a= 8. Solve the triangle.
• First, you have to draw a picture of the triangle described. (Make sure the letters are in alphabetical order going clockwise.
• Next, you would find leg c because it is opposite of the given angle, C.
• Plug into the formula: c^2= (5^2) + (8^2) – 2 (5) (8) cos (60 degrees)
• Take the square root of both sides (plug into calculator exactly as is): c= 7
• Now you would find angle A by plugging into the formula: 8^2= (5^2) + (7^2) – 2 (5) (7) cos A
• Cos A= ((8^2) - (5^2) – (7^2)) / ((-2 (5) (7)
• Take the inverse and plug it into your calculator: A= 81.787 degrees
• Add the two angles and subtract them from 180: B= 32.213 degrees

I feel bad for the swedish people who cant read my blog:/ so enjoy(:

Sååå idag söndag den 16 oktober. Idag kommer jag att lära / visa dig hur du använder seniS för Wal att lösa icke-rätt trianglesss!! yay. När du använder den omvända hittar du 2 svar (står så i noter). Du använder detta när du vet vinkeln på triangeln och det motsatta benet (mitt emot från det andra benet). Så nu im ska visa dig den formel som används för att lösa en triangel när man vet vinkeln på triangeln och det motsatta benet (mitt emot från det andra benet). Jag är bara ska visa dig formeln en gång så att du bättre att ägna stor uppmärksamhet åt denna rätt nowww!! Sina / a = sinB / b = sinc / c Och det är den formel som du kommer att använda för att lösa triangel när du vet vinkeln och det motsatta benet (mitt emot från det andra benet). Och nu när du vet den formel som används för att lösa triangel jag kommer, på djupet, visa dig ett exempel. Hitta en först. synd 30 grader / 4 = sin 135 grader / en (använd 30 graders vinkel, eftersom de redan ger dig Sidans längd som en) a = 4sin 135 grader / synden 30 grader a = 5,656 Nu hittar vi b. synd 30 grader / 4 = sin 15 grader / c C = 4sin 15 grader / synden 30 grader C = 2,072

seniS fo waL 3-9

Soooo today is Sunday October 16. today i will be teaching/showing you how to use seniS fo waL to solve non-right trianglesss!!!! yay. When you use the inverse you will find 2 answers(says so in notes). You use this when you know the angle of the triangle and the opposite leg(opposite from the other leg). So now im going to show you the formula used to solve a triangle when you know the angle of the triangle and the opposite leg( opposite from the other leg). I'm only going to show you the formula once so you better be paying close attention to this right nowww!!!!

sinA/a=sinB/b=sinC/c

And thats the formula that your going to be using to solve the triangle when you know the angle and the opposite leg(opposite from the other leg).

And now that you know the formula used to solve the triangle I will,in depth, show you an example.

Find a first.

1. sin 30 degrees/4=sin 135 degrees/a (use the 30 degree angle because they already give you the side length for that one)
2. a=4sin 135 degrees/sin 30 degrees
3. a=5.656

Now we will find b.

1. sin 30 degrees/4=sin 15 degrees/c
2. c=4sin 15 degrees/sin 30 degrees
3. c=2.072

Law of Cosines

Law of Cosines is your last result for solving a triangle. You use it when Law of Sines doesn't work.

Formula for Law of Cosines: Opposite leg^2=adjacent leg^2+other adjacent leg^2=-2(adjacent leg)(other adjacent leg)cos(angle between the two adjacent legs

Ex.1 In triangle ABC, angle B=130, c=5, and a=8. Find angle A, angle C, and leg b.

• First, you need to draw a picture of the triangle:
• Second, you want to find leg b since it is the opposite of angle B.
• You plug the numbers into the formula: b^2=5^2+8^2-2(5)(8)cos130. Now you take the square root of both sides. When you plug that into your calculator, you get 11.850. So b=11.850.
• Now you can find angle A by using the formula. 8^2=5^2+11.850^2-2(5)(11.850)cosA. Then you subtract 5^2+11.850^2 to the other side, giving you 8^2-5^2-11.850^2=-2(5)(11.850)cosA. Now take the inverse of cosA. A=cos-1((8^2-5^2-11.850^2)/(-2(5)(11.850))=31.142
• To find angle C, just subtract the angles from 180. 180-130-31.142=18.858.

-Amber :)

9-4 Law of Cosines :)

Law of Cosines is my favorite thing we've done so far in this chapter. It is used only when law of sines does not work.

There is just one formula that you have to use:

(Opposite leg)^2=(Adjacent leg)^2+(Other Adjacent leg)^2-2(Adjacent leg)(Other Adjacent Leg)COS(Angle Between)

Example 1

Solve the triangle:

a=8, b=5, C=60º

The first thing you need to do is draw a picture with what is given.

The second thing you need to solve for is c.

Using the formula, you get c^2=5^2+8^2-2(5)(8)Cos 60º

Solv for c, c=SquareRoot(5^2+8^2-2(5)(8)Cos 60º)

c=7

Now that you have c, you need to solve for the angles. I'm going to solve for angle A first.

Using the formula, you get 8^2=7^2+5^2-2(7)(5)CosA

A=Cos^-1((8^2-7^2-5^2)/(-2(7)(5)))

A=81.787º

Now to find angle B,

180º-81.787º-60º=38.213º

B=38.213º

----Carleyyy!

9-1

9-1 Solving Right Triangles

In order to solve a right triangle, we use SOHCAHTOA:
sine(theta)=opposite/hypotenuse cosine(theta)=adjacent/hypotenuse tangent(theta)=opposite/adjacent
You can only use these formulas for right triangles. You can use any angle except the 90 degree.

Example 1:

• cosine 25 degrees=c/8


• c=16.314


• sin 25 degrees=18/b


• b=24.267

Example 2:

• cos 37 degrees=25/x


• x=31.303


• tan 37 degrees=y/25


• y=18.839

Example 3: Find the measures of the acute angles of a 3-4-5 right triangle.

• sin x=3/5


• x=sin^-1(3/5)


• =36.861 degrees


• tan x=4/3


• x=tan^-1(4/3)


• =53.130 degrees

Example 4: The legs of an isosceles triangle are each 21 cm long and the angle between them has measure 52 degrees. What is the length of the third side?

• sin 26 degrees=x/21


• x=9.206 cm


• 9.206(2)


• =18.412 cm

Example 5: In Triangle ABC,

• sin 25 degrees=b/18


• b=7.607


• cos 25 degrees=c/18


• c=16.314

-Jordan Duhon

Law of Cosines

Theres a triangle with legs 4, 8, 10 labeled ABC. a = 4, b = 8, c = 10. Find all of the angles.

Formula: opposite side^2 = adj. side^2 + other adj. side^2 -2(adj. side)(other adj. side)cos(angle b/w)

It doesn't matter which angle u solve for first but I'm going to solve for A first.

4^2 = 8^2 + 10^2 -2(8)(10)cosA (You don't have an angle so you will eventually have to take the inverse).

(Subtract 8^2 + 10^2)

4^2-8^2-10^2=-2(8)(10)cosA (Divide)

A= cos(inverse) ((4^2-8^2-10^2)

                            (-2(8)(10))) 

=  22.332 (You enter that into your calculator exactly) (Label angle A on your triangle)

You would do the same thing for the next angle (I'm solving for B)

You would get Angle B = 49. 458 and label your triangle

Add the two angles you found and subtract that number from 180 to find Angle C.

Angle C = 108.210

~ Parrish Masters

9-3 Law of Sines!

This week I am going to explain how to solve triangles using the law of sines.

This is only used when you know an angle and an opposite leg. This is also only used on non-right triangles.

The formula for these types of problems is:

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

Note: To solve this formula, you are going to cross multiply.

Now I am going to work a few examples of these types of problems.

Example 1:

You are going to solve for b and c. To do this you will use the formula given above. Lets solve for b first.

We are going to use Angle A to help solve for both side lengths because that side length is already given.

1. Sin 45 degrees/14=Sin 60 degrees/b
2. bSin 45 degrees=14Sin 60 degrees
3. b=14Sin 60 degrees/Sin 45 degrees
4. b=17.149

Now we are going to solve for c.

1. Sin 45 degrees/14=Sin 75 degrees/c
2. cSin 45 degrees=14 Sin 75 degrees
3. c=14Sin 75 degrees/Sin 45 degrees
4. c=19.129

Example 2:

You are going to solve for a and c. We are going to use Angle B because the side length is already given for that one. We are going to solve for a first.

1. Sin 30 degrees/9=Sin 105 degrees/a
2. a=Sin 30 degrees=9Sin 105 degrees
3. a=9Sin 105 degrees/Sin 30 degrees
4. a=17.388

Now lets solve for c.

1. Sin 30 degrees/9=Sin 45 degrees/c
2. cSin 30 degrees=9Sin 45 degrees
3. c=9Sin 45 degrees/Sin 30 degrees
4. c=12.726

And that is how you use Law of Sines to solve for triangles!

--Halie! :)

9-3 law of sines

today i am going to teach you the law of sines and how to solve a triangle with these laws. this can only be used on non right triangles. you should usually get 2 answers for the answer. it is only used when you know an angle and its opposite leg. the formula for it is sin a/a = sin b/b = sin c/c. you also cross multiply to solve. here is an example to help you out.



EX: angle A= 45 degrees  angle B= 60 degrees   a= 14

SOHCAHTOA

We spent this entire week reviewing for an taking tests except for Friday so SOHCAHTOA it is.

sinƟ=opposite/hypotenuse

cosƟ=adjacent/hypotenuse

tanƟ=opposite/adjacent

-This is only used on RIGHT triangles. A right triangle has one right angle.

-You CANNOT use the 90° angle in these types of problems.

-An angle of elevation

-An angle of depression

-When labeling your triangle you must do so alphabetically clockwise.

EXAMPLE: A 25 ft. pole is leaning against a building at an inline of 75°. Find the distance from the base of the building to the foot of the pole and the height of the pole.

First, draw yourself a triangle and label it.

I’m going to solve for x using cosine.

Cos75°=x/25 (using calculator)

X=6.470, add this to the sketch of your triangle ( I will show you the final drawing at the end of this blog)

Next, I’m going to solve for y using sine.

Sin75°=y/25 (using calculator)

Y=24.148, add this to the sketch also.

Your final sketch should look like this (:

There you have it. Oh and good luck with everyone’s exams (:

--Sarah

CRAP!!! WAIT, on my first blog, you should do tan38=13/9 and cross multiply! IM STUPID, DONT LOOK AT THIS B-ROB!!! IM LOSING IT BUT I CAUGHT MY MISTAKE YOU BETTER NOT TAKE OFF POINTS I DONT HAVE MY BINDER

So i jsut wrote a whole episode of the "Cory Teaches You How to do Something Show" and i left the page on accident and lost all my work. Im just thrilled right now, lemme tell ya! So i hate to do this, but im cancelling this weeks episode and doing a regular blog for once. I apologize for this but i have bio 2 to study for and i dont have time to do it all over again.

This week we learned about finding the length and angles of triangles. The first thing to do is learn SOHCAHTOA. This stands for:

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

Example: Using the picture above, say B= 38degrees a= 18 b= 26 Fnd A and c

First, lets find c. Do tan= opposite/ adjacent =26/18= 13/9

c=13/9

A= 180-90-38=52 degrees

I hope this is right because i have so much crap to do tonight for homework. Goodnight everybody!!

9-1

sooo today im going to be doing my blog on 9-1 which is going to help you out alot when you are looking for the side lengths of a triangle. your going to use sin cos and tan.

to help remember the equations for this process you can use SOHCAHTOA.

sinx=opposite/hypotenuse(or however you spell it)

cosx=adjacent/hypotenuse(or however you spell it)

tanx=opposite/adjacent.

you can only use these formulas to find the side lengths of right triangles

you cant use the 90degree angle

now im going to do an example of this

Example: Find x and y

So now your going to solve for x and y. Your going to use the 37 degree angle. You will use the formula cosx=adjacent/hypotenuse, which is cos37=25/x, which is going to equal 31.303 degrees.

Then to solve for y you will use tanx=opposite/adjacent, which is tan37=y/25, which is going to equal 18.839 degrees.

Your answers are going to be x=31.303 degrees and y=18.839 degrees.

9-1

This week I am going to teach you how to use sine, cosine, and tangent to find side the lengths and angle measures of a right triangle. This is a very easy thing to do.
There are three formulas that help us to do this:
• Sin x = opposite / hypotenuse
• Cos x = adjacent / hypotenuse
• Tan x = opposite / adjacent
The word SOHCATOA is used to help us remember these three formulas.
These formulas will ONLY work on RIGHT triangles. Whenever you find the side lengths, you NEVER use the 90 degree angle.
Now it is time to show you an example.
EXAMPLE: Find x and y (picture at top)
First we are going to find side length x. You are given the angle 35 degrees and the hypotenuse, 7. You would use sin, which gives you sin 35 degrees=x/7. You would then multiply both sides by 7, which gives you 4.015. So, x=4.015.
Next we are going to find side length y. This time you would use cosine, which gives you cos 35 degrees=y/7. When you multiply both sides by 7, it gives you y=5.734.

Section 9-1

So today we are going to be learning some geometry. This stuff is super easy because it's basically just putting things into your calculator. You must learn SOHCAHTOA. This means that sine=Opposite/Hypotenuse, Cosine=adjacent/Hypotenuse, and Tangent=Opposite/Adjacent.

When finding an angle, you take the inverse of either Sine, Cosine, or Tangent.

Now for the example:

Let's say that:

A=32º

a=42

b=48

Find B and c

The first we are going to find B.

So you use the TOA part of SOHCAHTOA.

You are finding the angle, so you get Tan^-1=48/42

Which gives you 48.814

Next we are going to find c.

so you use the SOH par of SOHCAHTOA

Sine32º=42/c

Solve for c so you get 42/Sin32º

Which gives you 79.257

So you're final answers would be B=48.813 and c=79.257º

Hope you learned something!

---Carleyyy!

Section 8-5 Solving More Difficult Trig. Functions

sec^2 x = 9
These trig functions are solved very similar to way they were done in Section 8-1
This time you would take the square root of 9 and get 3.
Then take the inverse of sin3 in you calculator and get no solution because Sin can not be greater than 1.
That would make your final answer no solution.

- Parrish Masters

9-1

This week I'm going to explain to you how to use sinx, cosx, and tanx to find side lengths of a triangle.

To remember how to do this you are going to use the word SOHCAHTOA. This is an easy way to remember the formulas to do this.

• Sinx=Opp/Hyp
• Cosx=Adj/Hyp
• Tanx=Opp/Adj

These formulas are only to be used of RIGHT triangles. Also, when finding these side lengths never use the 90 degree angle. Always use the other angle given.

Okay, so now I'm going to do an example.

Example 1: Find d and f

Okay, so we are going to find the side length f first.

Since you are give angle E, which is 12 degrees and you need to find f. You are going to use tanx=opp/adj. Which is going to give you tan12=9/f. When you solve for f you are going to get f=42.342 degrees.

Now we are going to find the side length d.

Your going to use angle E again, but this time you are going to use sinx=opp/hyp. This is going to give you sin12=9/d. When you solve for d you are going to get d=43.288 degrees.

So you answers are going to be f=42.342 degrees and d=43.288 degrees.

--Halie!

8-5

8-5 is almost exactly like 8-1.
***You can not divide by a trig function to cancel it on both sides in an equation. You must move it to the other side and factor it out.***
I will tell/show you the steps as I work the problem.

Ex.1: 8sin^2x-4=0

• First, you have to get the trig function by itself. You subtract 4 on both sides leaving you with 8sin^2x=4.
• Now you have to divide by 8 on both sides leaving you with sin^2x=4/8 which equals to sin^2x=1/2.
• After you do that, you have to take the square root on both sides. That gives you sinx=the square root of 1/2 which equals to sinx=+-square root of 2/2.
• Now you solve for x which gives you x=sin-1(+-square root of 2/2). When you are solving for x, you take the inverse of sin.
• Now you have to find what quadrant it is in. Sin-1(+-square root of 2/2)=45. So 45 is in quadrant one and since it is a positive and negative answer, you have to find all of the quadrants.
• Quadrant 2: -45+180=135. Quadrant 3: 45+180=225. Quadrant 4: -45+360=315. Your final answer: X=45 degrees, 135 degrees, 225 degrees, 315 degrees.

Ex.2: 6sinx=cosx for 0 degrees less than or equal to x less than or equal to 360 degrees

• First, you have to divide by cosx on both sides leaving you with 6sinx/cosx=1.
• Now you can change 6sinx/cosx to 6tanx so that gives you 6tanx=1.
• Then you have to divide by 6 on both sides which gives you tanx=1/6.
• Now solve for x: x=tan-1(1/6). tan-1(1/6)=9.462 which is in quadrant one.
• Now you have to find the quadrants where tan is positive. Tan is positive in one and three. We already found quadrant one so now we have to find quadrant three. Quadrant three: 9.462+180=189.462.
• Now you have to put those answers into degrees minutes and seconds. Quadrant one: .462x60=27.72, .72x60=43.2. Quadrant three: .462x60=27.72, .72x60=43.2.
• Final answer: X= 9 degrees 27 minutes 43 seconds, 189 degrees 27 minutes 43 seconds

Amber :)

8-4

8-4 Relationships Among the Functions

Today I'm doing my blog on 8-4 because I can guarantee you that I bombed this stuff on my test, and I need all the practice I can get. The following are all the relationship equations you will use in solving the problems in this section.

-sin x/cos x= tan x cos x/sin x= cot x

-Reciprocal Relationships:
csc(theta)=1/sin(theta) sec(theta)=1/cos(theta) cot(theta)=1/tan(theta)

-Pythagorean Relationships:
sin^2(theta)+cos^2(theta)=1 1+tan^2(theta)=sec^2(theta) 1+cot^2(theta)=csc^2(theta)

-Cofunction Relationships:
sin(theta)=cos(90-theta) & cos(theta)=sin(90-theta)
tan(theta)=csc(90-theta) & cot(theta)=tan(90-theta)
sec(theta)=csc(90-theta) & csc(theta)=sec(90-theta)

Example 1:
sec x - sin x tan x
=1/cos x - sin x (sin x/cos x)
=1-sin^2 x/cos x
=cos^2 x/cos x
=cos x
answer= cos x

Example 2:
Prove cot A(1+tan^2 A)/tan A= csc^2 A
=(cot A) (sec^2 A)/tan A
=(cos^2 A/sin^2 A) (1/cos^2 A)
=1/sin^2 A
=csc^2 A
answer= csc^2 A

*Note that when you're dealing with negatives in these problems, treat them just as they'd be positive, using all of your formulas.

-Jordan Duhon

review on 8-5

today i am reviewing 8-5 because this week was mostly review. rememebr that the same rules in 8-1 apply to the rules of 8-5 except these problems are harder. also the major rule is that you cannot divide by a trig function to cancel it on both sides in an equation. you must nove it to the other side and factor it out.



EX: 2 tan^2 x = 3 tan x - 1
 first move 3 tan x - 1 to the other side because you cannot divide by a trig function to cancel.
2 tan^2 x - 3 tan x + 1 = 0
factor: 2 tan^2 x - 2 tan x - 1 tan x + 1
group: (2 tan^2 x - 2 tan x) (-1 tan x + 1)
simplify: 2 tan x(tan x - 1)   -1(tan x - 1)
set to zero: 2 tan x - 1 = 0      tan x - 1 = 0
do the simple equation: tan x = 1/2     tan x = 1
inverse: x = inverse of tan (1/2)        x = inverse of tan (1)

Pythagorean Identites

1+tan^2x=secx

1+cot^2x=cosx

sin^2x+cos^2x=1

These pythagorean identites are the gist of how to figure out if there are trig identities in your problem. Knowing these identities is essential to one's success in getting to the right answer.

For example in verifying identites:

1. cos^2x-sin^2x =1-2sin^2x

Start with left side using the identity sin^2x+cos^2x=1 or cos^2x= -sin^2x+1

2. cos^2x-sin^2x= (1-sin^2x)- sin^2x

3. It equals 1-2sin^2x

8-4 (and no bad algebra)

This week we learned about relationships among trig functions…how riveting. The first thing you will need to know is your Pythagorean Identities.

Sin^2Ɵ+cos^2Ɵ=1, tan^2Ɵ+1=sec^2Ɵ, cot^2Ɵ=csc^2Ɵ

Now, the following steps are to be loosely followed.

1. Algebra (In this step you should factor, FOIL, etc. oh and don’t do it badly.)

2. Identities (first you will want to try one of the Pythagoreans I listed earlier. If none of these work move everything to sin and cos if that helps)

3. ALGEBRA :D but don’t do it badly.

4. Continue with steps 1-3, but no bad algebra.

It may also help you to look back over some algebra to save yourself some time while doing these sorts of problems.

Now, time for the example :D

Solve.

(sinƟ-1)(sinƟ+1)

The first thing you want to do here is FOIL (without bad algebra).

Sin^2Ɵ-1

Next, you’ll want to check your Pythagorean identities and, lucky for me, this is one! In the margin, write that sin^2Ɵ+cos^2Ɵ=1

Subtract sin^2Ɵ from both sides and you will discover your answer is cos^2Ɵ.

Oh and don't forget, no bad algebra.

--Sarah

Relationships among the functions

Soo today we will be learning about relationships among the different functions such as cos tan and sin. Next i will explain in deep details how you will deal with these functions.

-The first step is to use your algebra skills to add,subtract,multiply,or divide any of the fucntions that can be simplified.

-The next step is to use your identities to simplify.

-Next you will do algebra again to try to simplify more.

-You will repeat the same steps over and over again till your equation is in simplest form.

PYTHAGOREAN

sin^2x+cos^2x=1

1+cot^2x=csc^2x

1+tan^2x=sec^2x

Recipricles

cscx=1/sinx

tanx=sinx/cosx

secx=1/cosx

cotx=cosx/sinx

So now that i have explained to you how to solve/simplify these function and showed you the identities I will show you how to work them.

Examples:

(sec^2x-1)(csc^2x-1)

(tan^2x)(cot^2x)

(sin^2x/cos^2x)(cos^2x/sin^2x)

0

1+cot^2a reduces to 1/sin^2a