### (>O_o)> .:Vectors:. <(o_O<)

Hi everyone, today we did a lot of stuff in our class because we had two periods of math today, so let's get it started. At the beginning of the first period class, Mr. K gave us some problems to solve and he also reminded us about SOHCAHTOA. SOHCAHTOA means Sin= O/H, Cos= A/H, Tan= O/A. O means the opposite, A is the Adjacent ( Next to; neighboring) and H for Hypotenuse (Note: The hypotenuse is the longest side of a right triangle).

*Sin^-1, Cos^-1, and Tan^-1 are called

Arc Sin, Arc Cos, and Arc tan......This functions means the same thing.*

Here are the problems that Mr. K gave us:

1) Find: tan A, angle A, Length of AC

A) Tan A= O/A

Tan A = 5/12

Tan A = .4166

B) Angle A = arc tan (.4166)

Angle A = 22.61 degree

C) Length of AC

C^2= 25^2 + 5^2

C^2= 169 (Take the square root of C^2 and 169)

C = 13

2) Find: Sin R, Angle R, Length of QR

A) Sin R = O/H

Sin R = 11/25

Sin R = .44

B) Angle R= arc sin (.44)

Angle R= 26.10 degree

C) Length of QR

25^2= 11^2 + a^2

(25^2)-(11^2)= a^2

504 = a^2 ( Find the square root of 504 and a^2)

22.44 = a

3) Find : Length of XZ and length of XY

A) length of XZ

Cos 37= 32/H

H= 32 / Cos (37)

H= 40

B) length of XY

z^2 = x^2 + Y^2 - 2xyCos Z

z^2= (32^2 + 40^2) - (2*32*40*Cos (37))

z^2= 579.49 (square root both sides)

z= 24.1

4) Find: length of AC, Angle A

A) length of AC

b^2= a^2 + c^2 - 2*a*c*Cos B

b^2= (5.7^2 + 4.5^2) - (2*5.7*4.5*Cos (78))

b^2= 42.07 (square root both sides)

b= 6.49

B) Angle A

a/Sin A = b/SinB

5.7/Sin A = 6.49/Sin (78)

Sin A = (5.7*Sin (78))/6.49

Sin A = .8591

Arc sin (.8591)= 59.21 degree

5) Find: Angle C, b

A) Angle C

a / Sin A = c / Sin C

5.1 / Sin (28) = 3.8 / Sin C

Sin C = (3.8*Sin (28)) / 5.1

Sin C = .3498

arc Sin (.3498) = 20.47 degree

B) length of b

Step 1: Find Angle B so that we can use the sin law or the cos law.

Angle B = 180-28-20.47 = 131.53 degree

Step 2: Use sin law ( it's easier to work with :P)

a / Sin A = b / Sin B

5.1 / Sin (28) = b / Sin (131.53)

b= (5.1* Sin(131.53) / Sin (28)

b= 8.13

Mr. K also gave us two kinds of programs for our calculater. The programs is used to solve triangles, If you missed the class today be sure to ask someone to give you the program.

To access this programs:

1) Press the [PRGM] Key and then press [1]. The calculator will use the sin law and the cos law.

2) Press [PRGM] key and then press [2]. This will identify all the missing angles and sides.

For the second period of math, Mr. K gave us more problems to work on:

1) Hezy and Partick paddle their canoe 5.5 km west and then 4 km south. How far are they from their starting Point? What direction must they paddle to return to the starting point by the shortest route?

Therefore they are 6.8 km away from the starting point and must paddle north east to return to the starting point.

2) Forces of 210 N and 85 N act at an angle of 78 degree to each other. Find the resultant force and the angle it makes with the smaller force.

The resultant force is 242.38N, at 57.94 degree to 85N force. (sorry that I can't show you guys the solution but I'll have the solution by tomorrow).

Well today has been a crazy day and our homework is on page 322, numbers 1-4 and the next scribe will be (>O_o)> .: HEZY :. <(o_O<) .

*Sin^-1, Cos^-1, and Tan^-1 are called

Arc Sin, Arc Cos, and Arc tan......This functions means the same thing.*

Here are the problems that Mr. K gave us:

1) Find: tan A, angle A, Length of AC

A) Tan A= O/A

Tan A = 5/12

Tan A = .4166

B) Angle A = arc tan (.4166)

Angle A = 22.61 degree

C) Length of AC

C^2= 25^2 + 5^2

C^2= 169 (Take the square root of C^2 and 169)

C = 13

2) Find: Sin R, Angle R, Length of QR

A) Sin R = O/H

Sin R = 11/25

Sin R = .44

B) Angle R= arc sin (.44)

Angle R= 26.10 degree

C) Length of QR

25^2= 11^2 + a^2

(25^2)-(11^2)= a^2

504 = a^2 ( Find the square root of 504 and a^2)

22.44 = a

3) Find : Length of XZ and length of XY

A) length of XZ

Cos 37= 32/H

H= 32 / Cos (37)

H= 40

B) length of XY

z^2 = x^2 + Y^2 - 2xyCos Z

z^2= (32^2 + 40^2) - (2*32*40*Cos (37))

z^2= 579.49 (square root both sides)

z= 24.1

4) Find: length of AC, Angle A

A) length of AC

b^2= a^2 + c^2 - 2*a*c*Cos B

b^2= (5.7^2 + 4.5^2) - (2*5.7*4.5*Cos (78))

b^2= 42.07 (square root both sides)

b= 6.49

B) Angle A

a/Sin A = b/SinB

5.7/Sin A = 6.49/Sin (78)

Sin A = (5.7*Sin (78))/6.49

Sin A = .8591

Arc sin (.8591)= 59.21 degree

5) Find: Angle C, b

A) Angle C

a / Sin A = c / Sin C

5.1 / Sin (28) = 3.8 / Sin C

Sin C = (3.8*Sin (28)) / 5.1

Sin C = .3498

arc Sin (.3498) = 20.47 degree

B) length of b

Step 1: Find Angle B so that we can use the sin law or the cos law.

Angle B = 180-28-20.47 = 131.53 degree

Step 2: Use sin law ( it's easier to work with :P)

a / Sin A = b / Sin B

5.1 / Sin (28) = b / Sin (131.53)

b= (5.1* Sin(131.53) / Sin (28)

b= 8.13

Mr. K also gave us two kinds of programs for our calculater. The programs is used to solve triangles, If you missed the class today be sure to ask someone to give you the program.

To access this programs:

1) Press the [PRGM] Key and then press [1]. The calculator will use the sin law and the cos law.

2) Press [PRGM] key and then press [2]. This will identify all the missing angles and sides.

For the second period of math, Mr. K gave us more problems to work on:

1) Hezy and Partick paddle their canoe 5.5 km west and then 4 km south. How far are they from their starting Point? What direction must they paddle to return to the starting point by the shortest route?

Therefore they are 6.8 km away from the starting point and must paddle north east to return to the starting point.

2) Forces of 210 N and 85 N act at an angle of 78 degree to each other. Find the resultant force and the angle it makes with the smaller force.

The resultant force is 242.38N, at 57.94 degree to 85N force. (sorry that I can't show you guys the solution but I'll have the solution by tomorrow).

Well today has been a crazy day and our homework is on page 322, numbers 1-4 and the next scribe will be (>O_o)> .: HEZY :. <(o_O<) .

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