Q1 Find the length of 'x' in the figure below.

Note: The 1o` in triangle ADC wasn't given. We found that because we subtracted 30` from 40`.

IS X = a) 22 b)31 c)30 or d)52

The easiest way is to find the length of BC and BD using the trig. ratios (SOHCAHTOA) and then subtract BD from BC.Note: The 1o` in triangle ADC wasn't given. We found that because we subtracted 30` from 40`.

IS X = a) 22 b)31 c)30 or d)52

BC

tan(40) = BC / 200

BC = tan(40)*200

BC = 167.8

BD or h

tan(30) = BD / 200

BD = tan(30)*200

BD = 115.5

Therefore SINCE X=BC - BD

X=167.8 - 115.5

X=52.3

OR (d)

First off lets try to sketch the info given. tan(40) = BC / 200

BC = tan(40)*200

BC = 167.8

BD or h

tan(30) = BD / 200

BD = tan(30)*200

BD = 115.5

Therefore SINCE X=BC - BD

X=167.8 - 115.5

X=52.3

OR (d)

Q2 How many triangles can be constucted with A=30`, a=4, and b=10?

Is it: a) none b) 1 c) 2 d) Can't be determined?

Is it: a) none b) 1 c) 2 d) Can't be determined?

Using the sine law

Sin(30) / 4 = SinB / 10

SinB = Sin(30)*10 / 4

SinB = 0.5 * 10 / 4

SinB=1.25

ooooops!?! THE MAX VALUE FOR SINE OF AN ANGLE IS 1 (The # is too big!)

Q3 SOLVE 3CosA - 2 = 0 so that A is greater than 0` and less than 360`

3Cos(A) - 2 = 0

3Cos(A)=2

Cos(A)=2/3

A=Cos inverse (2/3)

A = 48`

But since Cos is positive in Quadrant 4 as well we have to get the other value that could possibly have a related angle of 48`

Therefore the other angle =360`- 48`= 312`

The answer to the Q is: A = 48`, 312`

Q4 How would you calculate the reference of the angle J pictured below

Refrence or R

R = 180 - J

and you are done!

Q5 AHN OBSERVES THA ANGLE OF ELEVATION OF A PLANE TO BE 48`. AT THE SAME INSTANT, THE ANGLE OF ELEVATION FOR CRYSTAL IS 34`. AHN AND CRYSTAL ARE 80KM APART ON LEVEL GROUND.

3Cos(A) - 2 = 0

3Cos(A)=2

Cos(A)=2/3

A=Cos inverse (2/3)

A = 48`

But since Cos is positive in Quadrant 4 as well we have to get the other value that could possibly have a related angle of 48`

Therefore the other angle =360`- 48`= 312`

The answer to the Q is: A = 48`, 312`

Q4 How would you calculate the reference of the angle J pictured below

Refrence or R

R = 180 - J

and you are done!

Q5 AHN OBSERVES THA ANGLE OF ELEVATION OF A PLANE TO BE 48`. AT THE SAME INSTANT, THE ANGLE OF ELEVATION FOR CRYSTAL IS 34`. AHN AND CRYSTAL ARE 80KM APART ON LEVEL GROUND.

A) How far is Ahn from the Plane?

B) How far is Crysta from the plane?

C) How high is the plane?

B) How far is Crysta from the plane?

C) How high is the plane?

Step1 sketch the info given.

There are 2 ways that this triangle.

Whereby P=Plane A=Ahn C=Crytal

Whereby P=Plane A=Ahn C=Crytal

Note: That the numbers in blue weren't given, but we can get the numbers if we solve the triangle using the 180` principle.

A)

Sin(34`)/AP = Sin(98`)/80

AP = 80Sin(34`)/Sin(98`)

AP=45Km

Sin(14)/80 = Sin(34)/AP

AP = 80Sin(34)/Sin(14)

AP=237.9KM

B)

Sin(48`)/CP = Sin(98`)/80

CP=80Sin(48`)/Sin 98

CP=60KM

Sin(14`)/80 = Sin(132`)/CP

CP = 80Sin(132`) / Sin(14`)

CP=245.7KM

C)

Using trig. ratios

SinC = h/CP

Sin(34`) = h/60

h = 60Sin(34`)

h = 33.6KM

Using trig. ratios again

SinA = h/AP

Sin(48`) = h/237.9

h = 237.9Sin(48`)

h=176.8KM

Sin(34`)/AP = Sin(98`)/80

AP = 80Sin(34`)/Sin(98`)

AP=45Km

Sin(14)/80 = Sin(34)/AP

AP = 80Sin(34)/Sin(14)

AP=237.9KM

B)

Sin(48`)/CP = Sin(98`)/80

CP=80Sin(48`)/Sin 98

CP=60KM

Sin(14`)/80 = Sin(132`)/CP

CP = 80Sin(132`) / Sin(14`)

CP=245.7KM

C)

Using trig. ratios

SinC = h/CP

Sin(34`) = h/60

h = 60Sin(34`)

h = 33.6KM

Using trig. ratios again

SinA = h/AP

Sin(48`) = h/237.9

h = 237.9Sin(48`)

h=176.8KM

I think I'm done.

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The next scribe will be.

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-Zeph

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The next scribe will be.

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-Zeph

I highly nominate this Scribe to be in the HALL OF FAME.

ReplyDeleteGreat use of pictures and colors.

AMazing concept

Great explanations

Overall a great piece of work