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TrigoNometry

Page history last edited by PBworks 17 years, 2 months ago

Learning Objectives

 

Problem 1 of 3

 

Tom and Jerry are 200 meters apart and on opposite sides of a flagpole. The angle of elevation from Tom to the top of the flagpole is 19°. The angle of elevation from Jerry to the top of the flagpole is 21°. How high is the flagpole?

 

Solution

 

Step One: The best thing to do first is to draw a diagram to go with the question

 

Step Three: The triangle now has 3 angles and 1 side. With this information we can use the Sine Law to solve for either side FJ or side FT.

|(200m)/(Sin140°) = (t)/(Sin19°)

t = 101.298827m|

 

Step Four: If you look at the triangle as two right angle triangles, we have just figured out the hypotenuse.

Using this right triangle we can then find out the third angle of this particular right triangle.

180° - 90° - 21° = 69°

 

Step Five: Now we have all the information we need to find the Height of the flagpole by using simple SOHCAHTOA.

|COS69° = (h)/(101.298827m)

h = (COS69°)(101.298827m)

h = 36.30225292m

h = 36.3 m|

 

Look at that! We've found the height of the flagpole.

 

Problem 2 of 3

 

A small lighthouse is on a man made island just offshore. The light can illuminate effectively up to a distance of 250 meters. From a point along the shore that is 500 meters from the lighthouse, the sightline to the lighthouse makes an angle of 20° with the shoreline. What length (to the nearest meter) of the shoreline is effectively illuminated by the light from the lighthouse?

 

 

Solution

 

Well since we know the lighthouse isn't on shore then we can assume that the length of shoreline illuminated by the light from the lighthouse is less than 500meters, because the light can only illuminate up to 250meters which is a lot smaller distance than 500meters.

To help solve the question it is easiest when we draw a diagram to help to visualize better what we are trying to achieve.

 

  • Well to start off ,we draw a diagram that would show where 250meters from the lighthouse would be.And you could also make the shortest distance from the lighthouse to the shoreline a right angle.

 

 

 

  • We could solve an angle using the sine law but we don't just want half of the distance illuminated by the lighthouse we want to whole thing. So we can make an addition to our first diagram by adding the exact same thing to the adjacent side seeing the whole length of illuminated shoreline

 

 

* Now that we do have the total distance that we want to find out,we can eliminate the stuff that we don't need, leaving the things needed to make our calculations.

 

  • With the information we have we can now use the sine law to find b in the largest triangle sin20°/250 =sin(b)°/500 cross multiplying 500sin20°/250= x

 

  • 500sin20°/250=.698 now we must do the inverse of .698 to find angle b .698sinˉ¹=44°

 

  • b=44° and since the triangle made with blue dotted lines is made from two lines that are 250meters, equally placed on either side of the right angle of the other triangle, both b's in the middle triangle are isosceles, making the b's the same degree. Both 44°

 

Using some fundamentals of geometry we also know that the sum of angles in any triangle is 180°

so a°+b°+c°=180°

44°+44°+c°=180°

c°=180°-44°-44°

c=92°

 

* Now that we have all the info we need we can once again use the sinelaw.

 

  • sin44°/250=sin92°/x
  • 250sin92°/sin44°=360meters

 

In conclusion the length of shoreline illuminated by the lighthouse is 360meters.

 

 

Problem 3 of 3

 

f(x) = -2sin(x - 45) + 1

 

(a) Sketch the graph of f(x) showing at least two cycles.

 

(b) Write a cosine equation, g(x), that is equivalent to f(x).

 

Solution

 

(a) Step 1

The first step is to graph y=sin(x)

 

 

Step 2

Graph the equation y=sin(x-45). It will shift the graph for equation y=sin(x) 45 units to the right.

 

Step 3

Graph the equation y=sin(x-45)+1. It will shift the graph for equation y=sin(x-45) one unit up.

 

Step 4

Graph the equation y=-2sin (x-45)+1. It will make the graph open down and have a maximum of two.

 

(b)Step 1

Since y=sin(x) and y=cos(x-90) are just the same.Shift the graph for the equation y=cos(x-90) 45 units to the right.

 

It will make the equation into y=cos(x-135)

 

Step 2

Shift the equation y=cos(x-135) one unit up.

 

It will make the equation into y=cos(x-135)+1

 

Step 3

Flip the graph to make it have a maximum of two.

 

IT will make the equation into y=-2cos(x-135)+1

Comments (1)

Anonymous said

at 1:48 am on Jan 28, 2007

Hey,
I edited your images so that the y-axis represents degrees, because if you look at what you did to the graph (shifted it 45 degrees to the right), you used degrees, so yea. Hope i didn't get you wrong.

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