10-2!

Formulas:

~tan(A+B)= (tanA + tanB)/(1-tanAtanB)

~tan(A-B)= (tanA- tanB)/(1+tanAtanB)

YOU MUST USE THE FORMULAS!

______________________________________________________________________________

Ex.1: Suppose tanA=2 and tanB=6; Find tan(A+B):

tan(A+B)=(2+6)/(1-2*6)

=8/1-12

tan(A+B)=-8/11

Ex.2: Suppose tanA=1/2 and tanB=3/4; Find tan(A-B):

tan(A-B)=(1/2-3/4)/(1+1/2*3/4)

=(-1/4)/(1+3/8)

-1/4/11/8= -8/44

tan(A-B)=-2/11

Ex.3: Suppose tanA=3 and tanB=5; Find tan(A-B):

tan(A-B)=(3-5)/(1+3*5)

=-2/1+15

=-2/16

tan(A-B)=-1/8

Ex.4: Suppose tanA=1/6 and tanB=2/3; Find tan(A+B):

tan(A+B)=(1/6+2/3)/(1-1/6*2/3)

=5/6/8/9

=45/48

tan(A+B)=15/16

10-1

cos(x+pie/2)= -sin x
cos x*cos pie/2-sin x*sin pie/2
0*cos x-sin x*1
= -sin x

To simplify these types problems, we are given six different formulas for this section.  The two main formulas we use for this section is cos(alpha+/- beta)=cos theta*cos beta-/+sin theta*sin beta and sin(theta+/- beta)=sin theta*cos beta+/- cos theta*sin beta.  For this problem, we had to show how cos(x+pie/2) is equal to -sin x.  To do this, you remember your formulas and see what replaces cos(x+pie/2) and that is cos x*cos pie/2-sin x*sin pie/2.  Cos pie/2 is on your trig chart and that equals 0.  Sin pie/2 is also on your trig chart and that equals 1.  Now you have cos x*0-sin x*1.  Cos x*0 equals 0, so -sin x*1 equals -sin x.  -sin x is your answer.

Addition formulas

These are forumulas used to simplify and evaluate trignomic equations. You are given instructions with a trig function and you plug into formulas. Sometimes you may need to find a reference angle before plugging in. Formulas can be used both backwards and like they are.
They can also be negative if necessary.

1.cos(@+B)=cos@ cosB + sin@ sinB

2.sin(@+B)=sin@ cosB + cos@ 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 sin X-Y/2

6.cosX-cosY = -2sin X+Y/2 sin X-Y/2

Ex1. sin (75)
a.Must first get two angels for formula: sin(45+30)
b.Then plug into formula: sin45 cos30 + cos45 sin30
c.Use trig chart to get value. (square root2/2) (square root 3/2) +
(square root2/2) (1/2)
=square root4/4

10-3 Double Angles and Half Angles

To find the what the double of sin or cos or whatever of an angle or half of sin or cos or whatever of an angle, you DO NOT multiply by two or divide by two. The sin(2a) or sin(a/2) is deceiving.

The formulas you'll need to know are:
sin(2a) = 2sin(a)cos(a)

cos(2a) = cos²(a)-sin²(a)

or = 1-2sin²(a)

or = 2cos²

tan(2a) = 2tan²(a)/1-tan²(a)

sin(a/2) = ±√(1-cos[a]/2)

cos(a/2) = ± √(1+cos[a]/2)

tan(a/2) = ± √(1-cos[a]/1+cos[a])

or = sin(a)/1+cos(a)

or = 1-cos(a)/sin(a)

Example:

If cosA = 1/3, find:

a. sin2A

b. cos2A

c. tan2A

d. sinA/2

e. cosA/2

f. tanA/2

First, we need to make a triangle to find the other trig functions.

cosA = 1/3, so 1 is x and 3 is the radius. We use the Pythagorean theorem to find y.
1²+y² = 3²

1+y² = 9

y² = 8

y = √(8) which can be simplified to 2√(2)

cosA= 1/3

sinA = 2√(2)/3

tanA = 2√(2)

a.sin2A = 2sinAcosA

= 2[2√(2)/3][1/3]

=4√(2)/9

b.cos2A = cos²A-sin²A
= (1/3)² - (2√(2)/3)²

= (1/9)-(8/9)

= -7/9

You could have used any of the three formulas for cos2A.

c. tan2A = 2tanA/1-tan²A

= 2[2√(2)]/1-[2√(2)]²

= 4√(2)/(1-8)

= -4√(2)/7

d.sinA/2 = ± √(1-cos[A]/2)

= ± √(1-(1/3)/2)

= ± √([2/3]/2)

= ± √(2/6)

= ± √(1/3)

e.cosA/2 = ± √(1+cosA/2)

= ± √(1+[1/3]/2)

= ± √([4/3]/2)

= ± √(4/6)

= ± √(2/3)

f.tanA/2 = sin(A)/1+cos(A)

= [2√(2)/3]/1+[1/3]

= [2√(2)/3]/[4/3]

= 6√(2)/12

=√(2)/2

Again, you could have used any of the three formulas.

10-3

10-3 has like a million formulas that we have to learn. So here they are:

1. sin2(alpha) = 2sin(alpha)cos(beta)

2. cos2(alpha) = cos^2(alpha)-sin^2(alpha)

3. cos2(alpha) = 1-2sin^2(alpha)

4. cos2(alpha) = 2cos^2(alpha)-1

5. tan2(alpha) = 2tan(alpha)/1-tan^2(alpha)

6.sin(alpha)/2 = +or- square root of 1-cos(alpha)/2

7. cos(alpha)/2 = +or- square root of 1+cos(alpha)/2

8. tan(alpha)/2 = +or- square root of 1-cos(alpha)/1+cos(alpha)

9. tan(alpha)/2 = sin(alpha)/1+cos(alpha)

10. tan(alpha)/2 = 1-cos(alpha)/sin(alpha)

Ex. 1: Simplify

2cos^2 10degrees-1

First, look at your formulas and see which one would fit the problem. You then see that you can use formula 4 and you replace the alphas with 10 degrees.

cos(2)(10) which is cos 20 degrees. Cos 20 degrees does not exist on the trig chat, therefore you cannot simplify it anymore.

Ex. 2: Simplify

2sin(alpha)/2cos(alpha)/2

First, you look to see what formula you would use for this problem. You can use formula 1. sin2(alpha/2). The twos cancel which leaves uou with sin(alpha)

Formulas for sin and cos

Yayy for no school tomorrow!
But we still have to blog.. :/
Haha, anywayysss.....

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

--Rewriting a Sum or Difference as a Product
sinx+siny=2sin((x+y)/2)cos((x-y)/2)
sinx-siny=2cos((x+y)/2)sin((x+y)/2)
cosx+cosy=2cos((x+y)/2)cos((x+y)/2)
cosx-cosy=-2sin((x+y)/2)sin((x-y)/2)

--The two main purposes for the addition formulas are:
1) To find the exact values of trigonometric expressions
2) Simplifying expressions to obtain other identities.

--The sum or difference formulas can be used to verify many identities that we have seen, and also to derive new identites.

^^that looks veryyyyy complicated, but it's not.

Example 1:
Simplify the given expression.
sin75degrees(cos15degrees)+cos75degrees(sin15degrees)
sin(75degrees+15degrees)=sin90degrees=1

Example 2:
Simplify the given expression.
cos5pi/12(cospi/12)-sin5pi/12(sinpi/12)
cos(5pi/12+pi/12)=cos6pi/12=cospi/2=0

Example 3:
Simplify the given expression.
sin3x(cos2x)-cos3x(sin2x)
sin(3x-2x)=sinx

Example 4:
Find the exact value of each expression.
cos105degrees
cos(60degrees+45degrees)=cos60degrees(cos45degrees)-sin60degrees(sin45degrees)
(1/2)(square root of 2/2)-(square root of 3/2)(square root of 2/2)
square root of 2/2-square root of 6/2=square root of 2-square root of 6/4

Example 5:
Simplify the given expression.
cos105degrees(cos15degrees)+sin105degrees(sin15degrees)
cos(105degrees-15degrees)=cos95degrees

Example 6:
Simplify the given expression.
sin4pi/3(cospi/3)-cos4pi/3(sinpi/3)
sin(4pi/3-pi/3)=sin5pi/3

Example 7:
Simplify the given expression.
cos2x(cosx)+sin(2x)sinx
cos(2x-x)=cosx

10-2

Sum formula for tangent: tan(α + β) = tanα + tanβ/1 – tanαtanβ

Difference formula for tangent: tan(α – β) = tanα - tanβ/1 + tanαtanβ

Ex 1. Suppose tanα = 1/3 and tanβ = ½

a. Find tan(α + β)

tanα + tanβ/1 – tanαtanβ

1/3 + ½ / 1 – (1/3)(1/2)

2/6 + 3/6 / 1 – 1/6

5/6 / 5/6 = 1

b. Find tan(α – β)

tanα - tanβ/1 + tanαtanβ

1/3 – ½ / 1 + (1/3)(1/2)

2/6 - 3/6 / 1 + 1/6

-1/6 / 7/6 = -6/42 = -1/7

10-3

In 10-3, like all the other lessons in chapter 10, there is a good bit of formulas.

1. sin2(alpha) = 2sin(alpha)cos(beta)
2. cos2(alpha) = cos^2(alpha)-sin^2(alpha)
3. cos2(alpha) = 1-2sin^2(alpha)
4. cos2(alpha) = 2cos^2(alpha)-1
5. tan2(alpha) = 2tan(alpha)/1-tan^2(alpha)
6.sin(alpha)/2 = +or- square root of 1-cos(alpha)/2
7. cos(alpha)/2 = +or- square root of 1+cos(alpha)/2
8. tan(alpha)/2 = +or- square root of 1-cos(alpha)/1+cos(alpha)
9. tan(alpha)/2 = sin(alpha)/1+cos(alpha)
10. tan(alpha)/2 = 1-cos(alpha)/sin(alpha)

Eample 1: Simplify the given expression
2cos^2 10degrees-1

So you look at your formulas to see which one you could use. You can use formula 4 and all you have to do is replace the alphas with 10 degrees.
cos(2(10)) = cos 20 degrees, this is not on the trig chart so you can not simplify it anymore.

Example 2: Simplify the given expression
2sin(alpha)/2cos(alpha)/2

You would use the 1st formula for this. sin2(alpha/2). The 2's cancel and you get sin(alpha)

Example 3: Find the exact value of each expression
2cos^2 3(alpha)-1

You use the 4th formula. cos2(3alpha) = cos6(alpha). That is your final answers

Example 4: Find the exact value of each expression
2tan(pi/8)/1-tan^2(pi/8)

You use the 5th formula. tan2(pi/8). You cange that to degrees pi/8 is 45/2 so when you simplify tht you get tan45 which when you look on the trig chart is 1

---Danielle :)

10-2

Formulas:

• tan (A+B) = ( tanA + tanB )/( 1 -tanAtanB )


• tan (A-B) = ( tanA -tanB )/( 1 + tanAtanB )

Ex1: If tan A = 1/4 and tan B = 1/5

a. Find tan (A+B)

(tan A + tan B )/( 1 –tan A*tan B ) is the formula. Now you need to REPLACE tan A and tan B for the values they gave you. YOU HAVE TO USE THE FORMULAS.

( 1/4 + 1/5 )/(1 -(1/4)(1/5)) You don't need a calculator to do this.

( 5/20 + 4/20 )/( (20/20) -(1/20))

(9/20)/(19/20)

180/380

9/19

b. Find tan (A-B)

the formula is ( tan A –tan B )/(1 + tan A*tan B). Now you have to REPLACE tan A and tan B for the values I gave you.

( 1/4 -1/5 )/( 1 + (1/4)(1/5))

( 5/20 -4/20 )/( (20/20) + (1/20))

( 1/20 )/( 19/20 )

20/380

1/19

Law of sines.

Last week, one of our lessons were Law of Sins
-You can only use Law of Sines in a non-right triangle when you have a pair, an angle and opposite side.

Formula:
sinA/a=sinB/b=sinC/c

Cross multiply to solve.

For inverses you get two answers which you then check to see if they work

Ex.1

ABC triangles:

angle C-25 degrees,side b-3,side c-2

First solve for angle B

Set up a proportion sin25/2 = sinB/3

cross multiply 3sin25/2 = 2sinB/2

SinB = -1(3sin25/2) = 39.34 degrees

Now that you have the angle measures to C and B, in order to find A you just subtract the sum of those two angles from 180 degrees.

For your second triangle,in order to find the degrees of angle B you have to make the one in the first triangle negetive and add 180.

After you do this,you will then be able to find your other angles.

To find the side lentgh of a you cross multiply

sin25/2 = sin115.66/a

cross multiply asin25/sin25 = 2sin115.66/sin25

a=4.265

Law of Sines

Used on in a non-right triangle when you have a "pair", which is a angle, and a opposite angle.

The formula is: Sin A/ a = Sin B/ b = Sin C/ c

Cross multiply to solve

You will get two answers for inverse

Check to see if they work or not

#Example 1

Solve

Sin 50/ 25 = Sin 110/a place

aSin50= 25 Sin110 cross multiply

a= 25 Sin 110/ Sin 50 = 30.667 plug into calc

Sin 50/25 = Sin20/c place

cSin50=25Sin20 cross multiply

c= 25Sin20/Sin50 = 11.162

law of sines!

During last week one of our lessons were Law of Sins
When using this,you can only use Law of Sines in a non-right triangle when you have a "pair", which is an angle and opposite side.

The Formula:
sinA/a=sinB/b=sinC/c

To solve the equation cross multiply

For inverses you get two answers which you then check to see if they work

Ex.1

ABC triangles:

angle C-25 degrees,side b-3,side c-2

First we are going to solve for angle B

In order to do this, set up a proportion sin25/2 = sinB/3

cross multiply 3sin25/2 = 2sinB/2

SinB = -1(3sin25/2) = 39.34 degrees

Now that you have the angle measures to C and B, in order to find A you just subtract the sum of those two angles from 180 degrees.

For your second triangle,in order to find the degrees of angle B you have to make the one in the first triangle negetive and add 180.

After you do this,you will then be able to find your other angles.

To find the side lentgh of a you cross multiply

sin25/2 = sin115.66/a

cross multiply asin25/sin25 = 2sin115.66/sin25

a=4.265

Area of a triangle

The area of a triangle can be found wheather it is a right triangle or not. There are two different formulas for right and non-right triangles. For a right triangle the formula is more simple and you do not need an angle since its obviously 90 degrees. With the formula for a non-right trianlge you need two legs and the angle inbetween them. If you dont have all of this given in the triangle, you must use other formulas such as law of sins and cos to find them. *Make sure after you solve the for the area you put the units after to revcieve full credit.

Formula for right triangle:

1/2 B H (base times height divided by 2)

Fromula for non-right triangle:

1/2 (adjacent leg) (adjacent leg) sin(angle in between them)

Ex1 ABC 1/2 *4 =4

Ex2 DEF 1/2 4*5*sin(50)=5

Area of a Triangle

You can find the area of any triangle if you are able to find two sides and the angle between the two sides.

The area of a triangle is equal to 1/2 times the base (or side 1), times the height (or side 2), times the sin of the angle between. Or, A=1/2(b)(h)(sin[angle])

If it is a right triangle, you can simplify it to A=1/2(b)(h) because sin(90) equals one.

Ex.1
Find the area of (in centimeters):

I drew a little dotted line where you should cut the figure so you can find the area. You don't want to cut it horizontally because that would destroy your right angle.

Now you have to find the area of each triangle you made and add them together.

The left triangle is a right triangle, and you can also see that it is a pythagorean triple. The missing side and dotted line is 13.

The area of the left triangle = 1/2(5)(12) = 30 cm^2.

The area of the triangle on the right is going to be a little more difficult to find. You have to find the angle of the left triangle to subtract to find the right.
We can use SOHCAHTOA, so the little part of 108 = sin^-1(5/13) = 22.620. Now we can subtract that from 108 to find the right side of 108.
108 - 22.620 = 85.38
Now we can use the formula to find the area of the triangle on the right.

The area of the triangle on the right = 1/2(13)(18)(sin85.38) = 116.620 cm^2.

Now you add them together 116.620 + 30 = 146.620 cm^2.

Law of Cosines

Used when Law of Sines doesn’t work

Formula: leg^2 = adj leg^2 + other adj leg^2 – 2( adj leg)(other adj) cos(angle between)

Ex 1. On triangle ABC side a =7, side c = 3, and angle B = 130 degrees. Find side b, angle A, and angle C

To find side b: b^2 = 3^2 + 7^2 -2(3)(7) cos130 degrees

Square root both sides and b = 9.219cm

7^2 = 3^2 + 9.219^2 – 2(3)(9.219)cosA

-2(3)(9.219)cosA = 7^2-3^2-9.219^2

cosA = ((7^2-3^2-9.219^2)/(-2(3)9.219))

A = cos^-1((7^2-3^2-9.219^2)/(-2(3)9.219))

Angle A = 35.574 degrees

130 + 35.574 = 165.574 – 180 = 14.426

Angle C = 14.426 degrees

9-2 AREA OF A TRIANGLE!!

The formula for area of a right triangle is Area = 1/2(base)(height)

The formula for area of a non-right triangle is Area = (adj. leg)(adj. leg)sin(angle between)

Example 1: In triangle ABC, a=2cm, b=11cm, and C=44° find the area.

Ok boys and girls, here you would use the non-right triangle formula because it's not right its not 90° not making it wrong or left its just special.

Area = 2cm*11cm*sin44°

Area=15.282cm²

Example 2: In triangle ABC, a=20miles, b=3miles, and C= 90°, find the area.

Ok this is a HUGE triangle, but here you would use the right triangle formula because it has a 90° angle.

Area= 1/2*base*height

Area = 1/2(20miles)(3miles)

Area = 30miles²

Example 3: In triangle QRS, r=5 milimeters, s=2.5 milimeters, and Q= 120°

Ok this is a baby one, but here you would use the non-right triangle formula because it doesn’t have a 90° angle

Area = 5 milimeters*2.5 milimeters *sin(120°)

Area = 10.825 milimeters²

9-2 Area of a triangle

For a right triangle use the formula: A=½bh

For a non-right triangle use the formula: A=½ (adj. leg) (adj. leg) (sin angle between)

Find the area of each triangle.

Ex. 1. a=4, b=5, angle C= 30°

For this problem you would use the formula A=½(adj. leg) (adj. leg) (sin angle between). So, A=½ (4) (5) (sin 30°) would give you the area of 5cm^2.

2. a=4, b=5, angle C=150°

For this problem you would use the formula A=½(adj. leg) (adj. leg) (sin angle between). So, A=½ (4) (5) (sin 150°) would give you the area of 5cm^2.

3. Find the area of triangle XYZ if x=16, y=25, and angle Z=52°.

For this problem you would use the formula A=½(adj. leg) (adj. leg) (sin angle between). So, A=½ (16) (25) (sin 52°) would give you the area of 157.602cm^2.

4. Find the area of triangle RST if angle S=125°, r=6, t=15.

For this problem you would use the formula A=½ (adj. leg) (adj. leg) (sin angle between). So, A=½ (15) (6) (sin 125°) would give you the area of 36.862cm^2.

Law of Cosines!

~Law of cosines is only used when law of sines doesn't work.

~The formula is (leg)^2=(adj leg)^2+(other adj leg)^2-2(adj leg)(other adj leg)cos(the angle in between)

___________________________________________________________________________________________________

~Ex.1: In triangle ABC: A=25, B=60, a=6, & c=8.

Find: C & b-

180-60-25=95

C=95

b^2=6^2+8^2-2(8)(6)cos(60)

*First you square root everything

*Then you plug it into your calculator exactly the way it is.

b=7.211

~Ex.2: In triangle XYZ: Z=43, X=16, x=4, & y=21

Find: Y & z-

180-43-16=121

Y=121

z^2=4^2+21^2-2(4)(21)cos(43)

*First you take the square root of everything

*Then you plug it into your calculator exactly the way it is.

z=18.279

~Ex.3: In triangle DEF: F=80, d=10, & e=3

FInd: D, E, & f-

f^2=3^2+10^2-2(3)(10)cos(80)

*First you take the square root of everything.

*Then you plug it into your calculator exactly the way it is.

f=9.929

10^2=3^2+9.929^2-2(3)(9.929)cosD

*First subtract the squares to the other side

*Next divide each side by -2(3)(9.929)

*Then take the inverse and plug it into your calculator

D=cos^-1((10^2-3^2-9.929^2)/(-2*3*9.929))

D=82.685

180-80-82.685=17.315

E=17.315

9-3

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

sin45/14=sin 60/b                                sin45/14=sin75/c                   180-45-60=75
b=17.146                                            c=19.124                               angle C=75 degrees

For law of sines, you have to have a pair to solve the triangle.  When you have a pair, you set it equal to the other side or angle you are given and solve.  In this problem, you are given the measure of angle B and angle A.  You are also given side a which is 14.  Since we are given both the angle and side A, then you use that as your pair.  You put sin of 45 over 14 equal to the other number you are given which is angle B.  You cross multiply and solve for side b.  Next you have to find your last angle, so you subtract the angles you already have by 180 and you get 75.  Then you have to find the last side of the triangle.  You use the pair you were given in the beginning which is sin 45/14 and set it equal to the angle and side you want to find which is c.  You cross multiply again and get that c is equal to 19.124.

Law of Cosines

The Law of Cosines is used only on non-right triangles.

The formula for the law of cosines is:

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

__________________________________________________________

Example:

Triangle ABC, where a = 7, c = 3, and B = 130 degrees.

b^2 = 3^2 +7^2 - 2(3)(7)cos130 degrees

b ~ 9.219

7^2 = 3^2 + 9.219^2 - 2(3)(9.219)cosA

-2(3)(9.219)cosA = 7^2 - 3^2 - 9.219^2

cosA = (7^2 - 3^2 - 9.219^2)/(-2(3)(9.219))

cosA = .813

There you go.

Law of Sines!

*I just got back from retreat and I am beyond tired, so im gonna try to hurry and finish this so I get some sleep before your test tomorrow..

--You can only use Law of Sines in a non-right triangle when you have a "pair", which is an angle and opposite side.

Formula:
sinA/a=sinB/b=sinC/c

--You cross multiply to solve.

--Two answers for inverse (you must check to see if they work).

Example 1:
Solve triangle ABC.
angleA=45degrees
angleB=60degrees
side a=14

sin45/14=sin60/b
bsin45=14sin60
divide both sides by sin45 to get b by itself
b=17.146
180-60-45=75degrees
angleC=75degrees
sin45/14=sin75/c
csin45=14sin75
divide both sides by sin45 to get c by itself
c=19.124degrees

Example 2:
Solve triangle ABC.
angleB=30degrees
angleA=135degrees
side b=4

sin30/4=sin135/a
asin30=4sin135
divide both sides by sin30 to get a by itself
a=5.657degrees
180-30-135=15
angleC=15degrees
sin30/4=sin15/c
csin30=4sin15
divide both sides by sin30 to get c by itself
c=2.07degrees

Example 3:
Solve triangle ABC.
angleC=25degrees
side b=3
side c=2

sin25/2=sinB/3
3sin25=2sinB
sinB=.634
B=sin^-1(.634)
B=39.346
39.346degrees and 140.564degrees
sin25/2=sin115.654/a
a=4.266degrees
180-39.346-25=115.654
180-140.564-25=14.436
sin25/2=sin14.436/a
a=1.180degrees

9-5 Course for a Ship

-Measure from North unless told otherwise


Example 1: A hunter walks east for hour and then north for 1 1/2 hours. What course should the hunter take to return to his starting point?



So you draw your triangle and discover that you need to figure out an angle that is missing. Still keeping in mind SOHCAHTOA you notice that you need to use tan because you have opposite over adjacent. Because you have to find the angle you need the inverse. x=tan-1(1/1.5) plug it into your calculator and your answer is x=33.690 degrees. What course should the hunter take to get back to his starting point? Your answer would be SouthWest.




Example 2: A sailboat leaves its dock and proceeds east for 2 miles. It then changes course to 205 degrees until it its due sourth of its dock. How far south is this?





So you need to draw a triangle again. You notice that you need to use SOCAHTOA once again. And again you need to use tan because your finding a side that you need opposite over adjacent. tan65=x/2 x=2tan65 and you plug that into your calculator and get x=4.289 miles.


I still don't really get this though...

--- Danielleeee

Zee Right Trianglee

Right Triangles

Trig relations can be used to find unknown parts of right triangles. 

                sinѲ=opposite/hypotenuse

                cosѲ=adjacent/hypotenuse

                tanѲ=opposite/adjacent

1)       In ∆ABC, angleA=90˚, angleB=25˚, and a=18.  Find b and c.

Side a is hypotenuse, side b is opposite, and c is adjacent.

Tan 25˚=b/18,  b=8.39

Cos25˚=c/18,  c=16.31

2)       In ∆DEF, angle D=90˚, angle E=12˚, and e=9.  Find d and f

Side e is opposite, side d is hypotenuse, and side f is adjacent.

Tan 12˚=9/f,  f=9/tan12˚, f=42.34

Sin 12˚=9/d, d=9/sin 12˚, d=43.29

There are reciprocal functions also

                secѲ=hypotenuse/adjacent

                cscѲ=hypotenuse/opposite

                cotѲ=adjacent/opposite

All right triangles can be solved with sin, cos, and tan but the reciprocal functions due exist.

Lovee,

Jenna Roussel

Law of Sines.

One of the 3 lessons that we've learned this past week is Law of Sines.

When using the Law of Sines, it is used on in a nonright triangle when you have a “pair”(angle of opp. Side).

The Formula is Sin A/a= Sin B/b = Sin C/c

When solving,you cross multiply.

For inverses you will get two answers,which you then have to see if they work.

Example : Solve each triangleABC.

Angle A= 58 degrees, Angle B= 65 degrees,and a = 11

The opposite pair is sin58/11.

In order to solve for little b, you must set up a proportion.

Sin58/11 = sin65/b

Cross multiply then you'll end up getting bsin58 = 11sin65

Divide each side by sin58 to get little b by itself and you then get b=11sin65/sin58

b equals 11.7556

To solve for little c, you set up sin58/11 = sin57/c

csin58=11sin57

Dive by sin 58

c=10.8784

Angle A= 58 degrees, Angle B= 65 degrees, C = 57 degrees, a = 11 b = 11.7556, and c =10.8784

Law of Sines :)

This week in class we learned about the law of sines. The law of sines are only used for non-right triangles when they have a pair, which is when you have the side and the angle opposite of it. You must fill out the triangle once you find the sides because if not you will get them all wrong because B-Rob is not searching through your work for your answer.

All triangles equal 180 degrees.

Formula: sin a/a = sin b/b = sin c/c

Cross multiply to solve!

You get two answers for inverses and you have to check them to see if they work.

Example 1.) Solve triangle ABC.

To find angle C you have to subtract 180-65-65 and get 50.

sin 65/6= sin 65/b

b=6

sin65/6=sin 50/c

c=5.071

Law of Sines

~This formula can only be used in non-right triangles.

~Can only be used when you have a "pair": an angle and and its opposite side

~Formula:

sinA/a=sinB/b=sinC/c

~You cross multiply to solve

~If you get an inverse, you must have two answers.

~Ex.1: Solve

A=40, a=6, B=60

~180-60-40=80

C=80

sin40/6=sin60/b

bsin40=6sin60

b=6sin60/sin40

b=8.084

sin40/6=sin80/c

csin40=6sin80

c=6sin80/sin40

c=9.193

~Ex.2: Solve

B=66, b=10, C=83

~180-66-83=31

A=31

sin66/10=sin83/c

csin66=10sin83

c=10sin83/sin66

c=10.865

sin66/10=sin31/a

asin66=10sin31

a=10sin31/sin66

a=5.638

9-4 law of cosines

angle C=60 degrees
side b=5
side a=8

c^2=5^2+8^2-2*5*8*cos60                                        
c=7

sin 60/7=sin A/8
8sin60=7sinA
A=sin^-1(8sin60/7)
angle A=81.787

180-60-81.787=38.213

First you have to know the formula for law of cosine. The formula is leg^2=adj. leg^2+other adj. leg^2-2*adj. leg*other leg*cos(angle between). Then you plug what you are given into to formula I just gave you to find one of the sides.  When you plug all that in you get that side c is 7.  Now you have to find the rest of the angles.  You can use law of sines like I did, or you can use law of cosine again.  I used law of sin and made a proportion thing.  I put the measure of angle C over side c equal to sin A over the measure of side a which is 8.  You have to take the inverse of sin because you are finding an angle now.  Angle A is 81.787, now you have to find the last angle of the triangle.  All you have to do now is subtract 180 from all the angles  you have to get 38.213.

law of sines

Waxxam wit it Mrs. Robinson, Ricky G reporting live from the Dooley mansion tryin not to get behind on these blogs again.... lets get it poppin

The formula for law of sines is sinA/a=sinB/b=sinC/c

ex: solve the triangle when A=40 degrees a=4 b= 3

1.) sin40/4=sin B/3

2.)cross multiply and get B= 3sin40/4

3.) B= .482

So basically you just take the given numbers and plug them into the equation
yeeaaaa there it is cuhh..... Rico Gatz signing out till next time ya digg

Law of Cosines

UUUGGGGHHHH this is my THIRD time writing this blog... I must be severely stupid to refresh the page TWICE...

Law of Cosines is used the find missing parts of a triangle when you aren't given opposite angles and sides. In other words, when you can't use Law of Sines or SOHCAHTOA. Law of Cosines is a last resort.

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

Ex. 1
b = 5, a = 7, and angle C = 40°. Solve the triangle. (This is the triangle above.)

Solving for c:
Plug into formula:
c^2 = 5^2 + 7^2 - 2(5)(7)cos(40)

Take square root:
c = sqrt[5^2 + 7^2 - 2(5)(7)cos(40)]

Plug into calculator:
c ≈ 4.514

Solving for angle A:
Plug into formula:
7^2 = 4.514^2 + 5^2 - 2(4.514)(5)cosA

Solve for cosA:
7^2 -4.514^2 -5^2 = -2(4.514)(5)cosA
(7^2 -4.514^2 -5^2) / (-2(4.514)(5)) = cosA
A = cos^-1[(7^2 -4.514^2 -5^2) / (-2(4.514)(5))]

Plug into calculator:
A ≈ 94.605°
≈ 83.395°

To solve for the angle of B, you can just subtract the angles from 180.
Scenario 1: 180 - 94.605 - 40 ≈ 45.395°

Scenario 2: 180 - 83.395 - 40 ≈ 56.605°


Scenario 1:
a = 7
b = 5
c = 4.514
angle A ≈ 94.605°
angle B ≈ 45.395°
angle C = 40°

Scenario 2:
a = 7
b = 5
c = 4.514
angle A ≈ 83.395°
angle B ≈ 56.605°
angle C = 40°

9-3 Law of Sines

You can only use this formula in a non-right triangle when you have an angle and opp. Side.

Formula: sinA/a= sinB/b= sinC/c

Steps: 1. Cross multiply to solve

2. two answers for inverse and you must check to see if they work

Ex. 1. angle A= 45°, angle B= 60°, a= 14

To find angle C subtract 180-45-60= 75°. To solve for b use sin45°/14= sin16°/b. Cross multiply to get bsin45°= 14sin60°. Which gives you b= 14sin60°/ sin45°. So b= 17.146. Then you must find c. To find c use sin45°/14= sin75°/c. Cross multiply to get csin45°= 14sin75°. c= 14sin75°/sin45°. c= 19.124.

2. angle B= 30°, angle A= 45°, b=9

To find angle C subtract 180-45-30=105. To solve for c use sin30°/9=sin45°/c. Cross multiply to get csin30°=9sin45°. Which gives you c= 9sin45°/sin30°. So c= 12.728. Then you must find a. To fin a use sin105°/a=sin30°/9. Cross multiply to get 9sin105°=asin30°. a=9sin105°/sin30°. a=17.387

Law of Sines

Used only in a non-right triangle when you have a “pair”(an angle and opposite side)

Formula: sinA/a = sinB/b = sinC/c

Cross multiply to solve

You get two answers for inverse and you have to check to see if they work.

Ex 1. Angle A = 110 degrees, Angle C = 20 degrees, and side b = 25. Find Angle B, side a, and side c.

180-110-20 = 50 degrees Angle B = 50 degrees

Sin50/25 = sin110/a sin50/25 = sin20/c

asin50 = 25sin110 csin50 = 25sin20

a = 25sin110/sin50 c = 25sin20/sin50

Side a = 30.667m side c = 11.162m

Law of Sines :)))

In C lass this last week one of the lessons learned was Law of Sines.

When using the Law of Sines, it is used on in a nonright triangle when you have a “pair”(angle of opp. Side)

The Formula is Sin A/a= Sin B/b = Sin C/c

When solving,you cross multiply.

For inverses you will get two answers,which you then have to see if they work

Example 1: Solve each triangleABC.

Angle A= 45 degrees, Angle B= 60 degrees,and a = 14

The opposite pair is sin45/14.

To solve for little b, you are going to set up a proportion.

Sin45/14 = sin60/b

After cross multiplying, you end up getting bsin24/sin45 = 14sin60/sin45

Divide each side by sin45 to get little b by itself and you then get b=14sin60/sin45

This equals 17.146

Now to solve for little c, you set up sin45/14 = sin75/c

csin45/sin45 = 14sin75/sin45

c=14sin75/sin45 =19.124

Angle A= 45 degrees, Angle B= 60 degrees, C = 75 degrees, a = 14 b = 17.146, and c =19.124

Law of Cosines

Wellllllll, here goes another blog... :)

It relates to the lengths of the sides of a triangle to the cosine of one of its angles.

The formula you use is:
c^2=a^2+b^2-2(a)(b)cosangle

Example 1:
angleC=60degrees
side b=5
side a=8

c^2=5^2+8^2-2(5)(8)cos60
*take the square root of everything
side c=7
sin60/7=sinB/5
*cross multiply
5sin60=7sinB
*divide both sides by 7
sinB=.619
*take the inverse
B=sin^-1(.619)
angleB=38.243degrees
*To find angleA, subtract 180-60-38.243
angleA=81.757degrees

Example 2:
angleR=120degrees
side s=14
side t=16

r^2=16^2+14^2-2(16)(14)cos120degrees
*take the square root of everything
side r=26
sin120/26=sinS/14
*cross multiply
14sin120=26sinS
*divide both sides by 26
sinS=.466
*take the inverse
S=sin^-1(.466)
angleS=27.775degrees
*to find angleT, subtract 180-120-27.775
angleT=32.225degrees

Example 3:
side z=41
side y=40
side x=9

40^2=9^2+41^2-2(9)(41)cosY
*Subtract 9^2+41^2 to the other side.
-2(9)(41)cosY=40^2-9^2-41^2
*Divide both sides by -2(9)(41) to get cosY by itself.
cosY=40^2-9^2-41^2/-2(9)(41)
*Take the inverse.
Y=cos^-1 40^2-9^2-41^2/-2(9)(41)
angleY=77.320degrees
41^2=9^2+40^2-2(9)(40)cosZ
*Subtract 9^2+40^2 to the other side.
41^2-9^2-40^2=-2(9)(40)cosZ
*Divide both sides by -2(9)(41) to get cosZ by itself.
41^2-9^2-40^2/-2(9)(40)
*Take the inverse.
Z=cos^-1 41^2-9^2-40^2/-2(9)(40)
angleZ=90degrees
angleX=180-77.320-90=12.68
angleX=12.68degrees

9-3 Law of Sines

-This week we learned law of sines. When learning law of sines you can only use in a non-right triangle when you have a "pair" (An anlge and opposite side).

-You crose multiply to slove and you will have two answers for inverse (must check to see if it works).

-Formula: sinA/a = sinB/b = sinC/c

Example 1:

The first thing we're gonna do is find little c. So you put sin45 over 14 equals sin75 over little c. Then you cross multiply and you get csin45 equals 14sin75. You need to solve for c so you divide by sin45 and you get c equals 14sin75/sin45. Plug that into you claculator and you get around 19.124

Now we're gonna solve for little b. You gonna put sin45 over 14 equals sin60 over little b. Then you will cross multiply again and you will get bsin45 equals 14sin60. You need to solve for little b so you will divide by sin45 and you will get b equals 14sin60/sin45. Plug into your calculator and you get around 17.146

Now to finish off the triangle you will need to find angleC so you subtract 180-60-45 and that gives you 75 degrees.

Fill in the rest of the triangle so Brob won't take off points and then your done.

---Danielle :)

Solving right triangles

To solve a right triangle you must use the rules of sohcahtoa. Of course sohcahtoa isnt a real word. Its abriviation for the rules to solve triangles. Note that these rules applies ONLY to right triangles. Also you cannont use the 90 degree angle in the formulas. You must use exsisting legs an angles to plug into formulas. You are ususally given three angels which always usually include a right angle. That leaves two angles for you to use. You must figure out how the angle and leg left are related then pick the formula that works best for it. Then you plug it into calulator. You must solve for all angles and legs to compelety solve a triangle. *To get full credit draw out triangles wheather its a words problem or not. Here are the rules to solve it:

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

Ex1 AB = 5 A=90 C=60
1. The other angle is 180-90-60=30. So B=30
2. Now use sohcahtoa 5 would be adjacent to angle B so lets use Tan
tan60=x/5 so when simplfied 5tan60=8.660
3.Now we must find other side. To use the same and lets use formula Cos
cos60=5/x simplified 5cos60=2.5
4.Now you have all sides and angles and are done.

LAW OF SINES

The law of sines is only used in a non-right triangle when you have a "pair" (an angle and it's opposite side)

FORMULA: (sinA / a) = (sinB / b) = (sinC / c)

*hint* cross multiply to solve*

You get two answers for inverses --> you must check to see if they work

Example 1: angle A = 52 degrees, a = 3, and angle C = 63 degrees. Find b, c, and B please.

Well since you asked nicely...

All triangles equal 180 degrees so...

180 -52 -63= B

B = 65 degrees

(sin 52) / 3 = (sin 65) / b

3 (sin 65) = b (sin52)

b=3 (sin 65) / (sin 52)

b = 3.45

(sin 52) / 3 = (sin 63) / c

3 (sin 63) = c (sin 52)

c = 3 (sin 63) / (sin 52)

c = 3.392

Example 2: a = 8, b = 2, c = 6, angle A = 100. Find B and C now!

(sin 100) / 8 = (sin B) / 2

2sin 100 = 8 sin B

sin B = 2(sin 100) / 8

B = arcsin ( 2(sin 100) / 8)

B = 14.253 degrees, 165.747 degrees but here 165.747 doesn’t work.

180 -100 -14.253 = C

C = 65.747 degrees

SOHCAHTOA

This week in class we learned about using SOHCAHTOA.

SOHCAHTOA stands for:

Sin=opposite/hypotenuse

Cos=adjacent/hypotenuse

Tan=opposite/adjacent

You can only use SOHCAHTOA when you are dealing with a right triangle.

You can not use the 90 degree angle.

The angle of elevation is when you are looking from the bottom to the top of the triangle.

The angle of depression is when you are looking from the top to the bottom of the triangle.

Steps:

Step 1: Find the names of the two sides you are working on. The side that you already know and the side that you are trying to find.

Step 2: Figure out which you will use.

Step 3: Fill in the equation with the information that you know.

Step 4: Solve the equation.

Example 1:

Find the height of the plane.

Step 1: The side that we know is the hypotenuse and the side that we are trying to find is adjacent to the angle.

Step 2: We are going to use Cosine.

Step 3: Cos60=h/1000

Cos60=0.5

0.5=h/1000
h / 1000 = 0.5
h = 0.5 x 1000 = 500

So, the height of the plane is 500.

Sectors of Circles

Yo Yo Yo you already know what time it is. its bout that time rico gatz gets his blog game finished. So check it out we gonna do some real shiznit with sectors of circles.

Formulas: k=1/2r^2($) k= area of sector r= radius $= central angle

k= 1/2 rs

Ex: A sector of a circle has arc length 4 cm and area 40m^2. FInd radius and Central Angle

1)Choose formula to plug in to: k=1/2 rs
2) 40=1/2 r(4)
.5 r=10
r=5
3.)40= .5(5)^2($)
4)$=40/12.5

$=3.2



SO there it is boi, rico gatz signing out for the night after that hard work of blogs i just put in. Time to chil for the night.... Ya Boi #21 out

Reference Angles

Wazzam you already know who it is, its ricky g in this blog putting in work ya feel me. Getting these blogs done for ya Mrs. Robinson fasho lets get it done

Let me show you what it do with reference angle real quick.
Steps: 1.) Sketch in original quadrant
2.) Positive or Negative with Unit circle
3.) Subtract 180 or 360 till between 0 and 90

Ex: Find reference angle of sin640

1) Its gonna be in quadrant 4
2) sin is negative
3) 640-360=280-360=-80

-sin80 is final answer

Yeah Mr. Beastly signing of but you already know ima be reporting live from the NYC