Thread: Problems need checking: Right Triangles, bearings,h. motion, & App. w/ Triangles

I have some problems, i have completed that needs to be checked.

Solving Right Triangles:

I'm given a right triangle where the outer points are a,b,c, and the angles inside the triangles, are A,B,C.

1. A = 41.5 degree, b = 20

20 tan (41.5 degrees) = 17.69 a

20/cos (41.5) = 26.70 c

48.5 B

2. A = 54.8 degrees, c = 80

35.2 degrees B

80 sin (54.8 ) = 65.37 b

6400-4225 = 21751 square equals 46.63 a


4. a = 11.2, c = 65.8


11.2 ^ 2 + b ^ 2 = 65. 8 ^ 2
4329.64
-125.44
-----------
4204.20 squared is 64.83 for b

sin-1(11.2/65.8) = 9.80 for A

80.2 degrees for B


5. 15.3 , b = 17.6

234.09 + 309.76 = 543.85 squared is 23.32 for c

tan -1 (17.6/15.3) = 49 degrees for A

41 degrees for B


6. b = 4, c = 9


9 ^ 2 - 4 ^ 2 =

81 - 16 = 65 a

cos -1 (4/9) = 63.61 degrees for A

26.39 degrees for B


Bearings:

N 60 degrees (B) W , N 75 degrees (A) E , W 80 degrees (C) S, S 35 degrees (D) E

Find the bearing from O to B (O is the origin)

N 30 degrees W

Find the bearing from O to D

S 55 degrees E

Harmonic Motion (Equations):

An object is attached to a coiled spring. The object is pulled down (neg. direction from the rest position) and then released. Write an equation for the distance of the object from its rest position after t seconds

8 inches ( Distnace ) 6 cent (Amp.) 4 (secs)

- 8 cos PI t


For this one, the object is propelled downward from its rest position at time t = 0. Write an equation for the distance of the object from its rest position after t seconds

Distance: 0 cent amp: 5 cent 2.5 secs.

* I'm not sure about my answer for B. This is what i came up with:

2 PI / 2.5 x 2/2 = 4 PI/5

Answer: 5 sin 4 PI /5


For these I'm given the equations, and i have to figure out the:

Displacement Max.

frequency

Time for one cycle.

1. d = 10 cos 2 PI t

Displacement: 10

freq. 2 PI / 2 PI x 1 / 2 PI = 1 per sec

1 sec (period)

2. d = - 8 cos PI/2 t

Displacement: 8

freq. PI / 2 / 2 PI x 1 /2 PI = 1 per sec


1 sec (period)



3. d = 1/3 sin 2 t

1/2 displacement

1/ 2 PI per sec

2 PI period



** I'm still working on some, so i'll try posting them up as i complete each section.

I did not go through all of them- but your answers for the first three are correct. (One point- in the first problem it would be better to specifically say "B= 48.5 degrees" rather than just "48.5 B"!)

Sorry bout that, i was kinda of in a rush. I'll post some of the word problems up in a bit.

Here are some of the word problems, some of them involve solving right triangles, finding bearings, and finally harmonic motion. Two of them i was not sure how to set them up so i didn't solve them yet, but i think i have an idea of how to set them up. Those are posted below as well.

From a point on level ground 30 yard from the base of a building, the angle of elevation is 38.7 degrees. Approximate the height of the building to the nearest foot.

30 x tan(38.7) = 24.03 ft


A 200 foot cliff drops vertically into the ocean. If the angle of elevation from a ship to the top of the cliff is 22.3 degrees, how far off shore, to the nearest foot, is the ship?

200/tan(22.3) = 487.65 ft


A police helicopter is flying at 800 feet. A stolen car is sighted at an angle of depression of 72 degrees. Find the distance of the stolen car, to the nearest foot, from a point directly below the helicopter.

Am i finding the base of the right triangle?


A building that is 250 feet high casts a shadow 40 feet long. Find the angle of elevation, to the nearest tenth of a degree, of the sun at this time.

tan-1(250/40)= 80.91 degrees

A flagpole is situated on top of a building. The angle of elevation from point on level ground 330 feet from the building to the top of the flagpole is 63 degrees. The angle of elevation from the same point to the bottom of the flagpole is 53 degrees. Find the height of the flagpole to the nearest tenth of a foot.

a) 330 x tan(53 degrees) = 437.92 ft

b) 300 x tan(63 degrees) = 588.78 ft

c) 588.8 - 437.9 = 150.9 ft

A boat leaves the enterance to a harbor and travels 40 miles on a bearing of S64 degrees. How man miles south and how man miles east from the harbor has the boat traveled?

a) 90 - 64 = 26 degrees

b) 40 cos (26 degrees) = 35.95 miles South

c) 40 sin(26 degrees) = 17.53

A ship sights a lighthouse direcly to the south. A second ship, 9 miles east of the first ship, als sights the lighthouse. The bearing from the second ship to the lighthouse is S 34 degrees W. How far to the nearest tenth of a male, is the first ship from the lighthouse?

a) 90 - 34 = 56 degrees

b) 9 x tan(56 degrees) = 13.34 miles

A ship is 9 miles east and 6 miles south of a harbor. What bearing should be taken to sail directly to the harbor?

a) 9^ 2 + 6 ^ 2 = 117 squared = 10.81 = 11 degrees

b) 90 - 11 = S 79 degrees W

A ship leaves port with a bearing S 40 degrees W. After traveling 7 miles, the ship turns 90 degrees and travels on a bearing of N 50 degrees W for 11 miles. At that time, what is the bearing of the ship from port?

I think for this problem 7 miles is the base of the right triangle but i'm unsure of where 11 miles goes. I know two triangles exist in this problem.

An object in simple harmonic motion has a frequency of 1/4 oscillation per minute and an amplitude of 8 feet. Write an equation in the form d = a sin w t for the object's simple harmonic motion.


6 sin 1/8 PI t