Trigonometry word problem!

[Maber]

we have a 10 worded problem about trigonometry and the last three are way different than the first seven, and I dont have a clue on how to even start with them

anyway here they are.

 

1.

Given a circle of radius 336.58 m, find the side of a regular circumscribed octagon

 

2.

A ship is sailing due east, parallel to the shore line, at 16 miles per hour. At 10 PM a light on shore is sighted due north. At 10:15 PM the lights has bearing N30°W. How far off shore is the ship?

 

3.

A boeing 777 aircraft takes off from NAIA on a runaway having a bearing 22°, after flying for 1 mile, the pilot of the aircraft request permission to turn 90° and head toward northwest, the request granted: a) what was the new bearing?, b. after the plane goes 2 miles in this direction, what bearing should the control tower use to locate the aircraft?

 

Hopefully someone can help with this I really dont know what to do 

[Nineshadow]

They are quite easy. Just think about it , draw how things should work.

[Maber]

I guess its easy but I dont really know what to do, can you give some clues ? 

[Philosobyte]

1.

• Each point on each side of the octagon must be connected to the center of the octagon by the radius of the circle. This forms a triangle. 
• The sum of the triangles' angles at the center of the circle must add up to 360 degrees. 
• You could use the law of cosines, or you could split the side at its midpoint to form two right triangles so you can use more basic trig.

2. 

• Construct a right triangle. If the ship is travelling 16 miles per hour, and 1/4 of an hour has passed, then how far has the ship traveled? That will be one of your sides. 
• The knowledge of that side, combined with the knowledge that the angle is 30 degrees, is all you need to complete the triangle and figure out all the other sides with trig functions.

3. 

• This problem is basically vectors. Part a should be easy. The plane is moving in the 1st quadrant of the graph. Turning 90 degrees northwest (counterclockwise) would mean you add the 90 degrees to the original angle. 
• For part b, you have two vectors, one of them (1, 22 degrees) and another of them (2, 112 degrees). Convert the vectors to complex rectangular form, add the vectors together, then convert them back to polar form to find the angle. Here's a resource about polar/rectangular conversion: http://www.teacherschoice.com.au/Maths_Library/Coordinates/polar_-_rectangular_conversion.htm

 

Feel free to ask for clarification.

Edited January 16, 2016 by Philosobyte

[Maber]

1.

• Each point on each side of the octagon must be connected to the center of the octagon by the radius of the circle. This forms a triangle. 
• The sum of the triangles' angles at the center of the circle must add up to 360 degrees. 
• You could use the law of cosines, or you could split the side at its midpoint to form two right triangles so you can use more basic trig.

2. 

• Construct a right triangle. If the ship is travelling 16 miles per hour, and 1/4 of an hour has passed, then how far has the ship traveled? That will be one of your sides. 
• The knowledge of that side, combined with the knowledge that the angle is 30 degrees, is all you need to complete the triangle and figure out all the other sides with trig functions.

3. 

• This problem is basically vectors. Part a should be easy. The plane is moving in the 1st quadrant of the graph. Turning 90 degrees northwest (counterclockwise) would mean you add the 90 degrees to the original angle. 
• For part b, you have two vectors, one of them (1, 22 degrees) and another of them (2, 112 degrees). Convert the vectors to complex rectangular form, add the vectors together, then convert them back to polar form to find the angle. Here's a resource about polar/rectangular conversion: http://www.teacherschoice.com.au/Maths_Library/Coordinates/polar_-_rectangular_conversion.htm

 

Feel free to ask for clarification.

 

I still cant understand no.1 but I managed to get an answer on no.2 which is 4.62 miles is this correct?, currently working on no.3 although I dont have any clue about vectors & polars yet. 

[Maber]

got an answer on no.3 as well, its (2.236,85°26'6'') is this correct? btw Im really thankful that site that you gave me is really easy to understand.

[Philosobyte]

1. Since you know that each vertex has a distance of 336.58 to the center of the circle (each vertex touches the circle) you can split the octagon into triangles. Since the sum of all the interior angles at the center of the octagon has to be 360 degrees, dividing them by 8 gives you 45 degrees for the innermost angle. Each triangle is isosceles, so you can split one of the triangles down the middle to give you two identical right triangles. Then you use the sine function to find the length of half a side; multiply that by two to get the length of the side: 257.6m. 

2. The picture is self-explanatory, I think, when you plug in the lengths and angles into the sine formula. The answer is 8 miles. What did you do differently? 

3. 85.43 degrees is the correct answer! congrats. I'm glad that you learned vectors and polar coordinates, but I thought of a much simpler method to solve this when I actually went to solve it *facepalm* Since you can form a right triangle with the sides 1 and 2, you can figure out the angle inside the triangle (63.43 degrees) and add 22 degrees to get the overall bearing.

[Maber]

OMG!, thank you so much for doing this I finally understand the first one.

 

and what I did in no.2 is I use sec 30° = x/4

divide both sides by 4 then multiply 4(sec 30°) and I got 4.618804 miles Im still confused on whether to use sin or cos.

 

Im really grateful for all of this really.

[Philosobyte]

OMG!, thank you so much for doing this I finally understand the first one.

 

and what I did in no.2 is I use sec 30° = x/4

divide both sides by 4 then multiply 4(sec 30°) and I got 4.618804 miles Im still confused on whether to use sin or cos.

 

Im really grateful for all of this really.

If you use secant, then it would be sec 30* = x/(unlabeled side)

and that's unsolvable because you have two variables.

 

Secant and cosine deal with the side which touches the angle, while cosecant and sine deal with the side opposite of the angle. The 4 in the triangle is on the opposite side, so you would have to use cosecant or sine. 

 

Another way to solve number 2 would be to use the 30 60 90 special right triangle. Any right triangle with the angle 30 must also have the angle 60, and vice versa. If the shortest side opposite 30 degrees is x, then the hypotenuse is 2x. Since in this case x = 4, the hypotenuse is 2(4) = 8.

(The leftover side is sqrt(3) * x, so in case you're wondering, the remaining side is 6.928). 

Glad I could help. 

[Maber]

I get it now, I just made a mistake in drawing the triangle and draw the angle on the wrong side thats why I use secant.

 

anyway you really helped me alot.