Vector Question Trigonometry

Vector questions trigonometry?

The boat's velocity will be
(30 km/h) i cos(theta) + (30 km/h) j sin(theta),
where theta is an angle measured CCW from due east.
The current's velocity is
(10 km/h) i cos(225 deg) + (10 km/h) j sin(225 deg),
and the resultant velocity is prescribed to be
(v) i cos(135 deg) + (v) j sin(135 deg).

Therefore you have
30 cos(theta) - 7.071 = -0.7071 v
30 sin (theta) - 7.071 = +0.7071 v
Adding these together gives
30 (cos(theta)+sin(theta)) = 14.142
30 cos(theta) = 14.142 - 30 sin(theta)
cos(theta) = 0.4714 - sin(theta)
cos^2(theta) = 0.22222 - 0.9428 sin(theta) + sin^2(theta)
1 - sin^2(theta) = 0.22222 - 0.9428 sin(theta) + sin^2(theta)
0 = 2 sin^2(theta) - 0.9428 sin(theta) - 0.77778
Let u = sin(theta)
0 = 2 u^2 - 0.9428 u - 0.77778
u = (0.9428/4) plus or minus (1/4)sqrt(0.9428^2+6.2222)
u = 0.2357 plus or minus (1/4)(2.66666)
u = 0.2357 plus or minus 0.6667
Since it is obvious that the boat must head in a generally northward direction,
the "minus" root is extraneous, and
sin(theta) = 0.2357 + 0.6667 = 0.9024, theta = 64.5 deg OR 115.5 deg,
and v = 30*0.9024/0.7071 - 10 = 28.3 (km/h)
To see which is right, go back to:
30 cos(theta) - 7.071 = -0.7071 v
Since 30 cos(64.5 deg) = 12.9 and 30 cos(115.5 deg) = -12.9,
it is evident that the right "v" will be obtained only if theta = 115.5 degrees.

(a) The boat must head 25.5 degrees west of north (or 64.5 degrees north of west).

(b) The boat's "ground" speed will be 28.3 km/h

Trigonometry: Vectors questions!?

1) vector v = < a , b > = < 50 cos π/4, 50 sin π/4 > + < 0 , -20>...|v| = ground speed , arctan (b / a ) = bearing {measuring from East}
2)The forces can be written as < 4.2 , 0 > and <10.3 cos 130°, 10.3 sin 130° > . Again v = their sum and arctan (b / a ) = angle measure {Θ} from the 1st force and 130 - Θ is the angle measure form the 2nd force..now do the computations

Vector Trigonometry Questions +10p?

I do not understand what "Letting be east and be north" means so I won't answer the questions directly. I will explain how I did it.

SW is 45 degrees east of south. Therefore Sin 45 or Cos of 45 x 11 = 7.78, these are the east and north components of 11 m/h SW.

I can now obtain the length of the actual vector by Pythagoras. (7.78^2 + 42^2)^1/2 = 42.71 mph

Its direction would be arctan (7.78/42) = 10.49 degrees east of south or accordingly to math rigor at
270 - 10.49 = 259.51 degrees of the circle.

Vector questions trigonometry?

Draw the vector triangle and mark all known angles

required track vector pointing NW ... [so that has to be the resultant course when the current and heading vectors are added]

from the TAIL of the required track vector draw the current vector pointing SW ... [b/c the current is coming FROM the NE so is actually going SW ... the opposite direction to NW] → mark it 10 km/h

then draw the heading vector from the head of the current vector to the head of the required track vector → mark it 30 km/h ... [the speed in still water means the speed the boat travels along the heading vector ... NOT the track vector]


The angle between the NW and SW vectors is 45 + 45 = 90° ... so the vector triangle is right angled and the simple trig ratios can be used

mark the angle between the hypotenuse (the heading vector) and the current vector θ

cos θ = adj/hyp = 10/30 = 1/3

so θ = arccos (1/3) = 70.5°

and 70.5° = ɸ + 45° ←←[where ɸ is the part of angle θ to the left of the vertical and b/c of alternate angles 45° is the part of angle θ to the right of the vertical]

so ɸ = 25.5° = 26° (to the nearest whole degree)

and the direction that the boat must head = 360 - 26 = 334° ... [and that can also be stated as N 26° W]



The speed that the boat actually travels over the ground is the speed it travels along the required track vector

so the speed at which the boat is traveling is the length of the required track vector

using the Pythagorean theorem:

speed over the ground = √[30² - 10²] = √[900 - 100] = √800 = √(400 * 2) = 20√2 km/h


so to summarize the answers:

a) to reach its destination the boat must head N 26° W ... [a bearing of 334°]

b) the boat will be traveling at 20√2 km/h

Finding Indicated Vectors (Trigonometry)?

Can someone help me solve this and explain it to me as well...
find the indicated vector let a=3i, b=i+j. find 6a+b

It's fairly easy but I am horrible at trig and I really need some help.

Thanks!

Trigonometry and Vectors help please?

doing thosee..

if you there can you plz write formulas... well ifs its not available that fine but its better...

well here what i got...
i used pythagorean theorem

1) A^2 + B^2 = C^2
(A = 160)^2 + (B = 90)^2 = (C)^2
(25600)+ (8100) = (C)^2
33700 = C^2 [add both values]
183.5755975068582 SE = C (NOT SURE THOUGH ARE YOU SURE ITS 90KM SOUTH WEST? OR SOUTH EAST... BECAUSE IT MAKE DIFFERENCE)

2) SAME FORMULA... this one for sure i got right....

10 blocks north ---> go up 10 make a line then right 10 beside it
15 block east ---> go right side 15 then right 15 beside it -- now tricky part comes...
11 block south ---> go down 11 which means subtract 10-11 which means there is no north.. so

final understanding is 1 block south and 15 block east...

A^2 + B^2 = C^2
(A = 1)^2 + (B = 15)^2 = (C)^2
(1)+ (225) = (C)^2
226 = C^2 [added both values]
15.033296378372908 SE = C [took square root out]

3) not sure but i will get back to you... i gotta refresh memory...

Need me = Ping me sorry for not getting answer right now..

Maths problem to do with vectors and trigonometry?

Find the length of the 2 Vectors U=3i+j+2k and V=-2i+j
Then find the dot product of U.V
Find the cosine of the angle between U and V.
Find the angle between the vectors.

Any feedback would be greatly appreciated!

Trigonometry Help on Vectors?

The missing thing in this problem is that a radio direction finder antenna reads the same from both sides. The peak reading is a straight line through the antenna which extends in both directions. If you draw the points A and B as the problem states and then draw the angles, then all you need to do is extend the line from B up and to the left until it intersects the line from A. That will be point C and the location of the transmitter. Now you have a triangle with angles 48 deg at A, 58 deg at B, and 74 deg at C.

The sine law states that:

b / sin B = c / sin C so:

b / sin 58 = 3.46 mi./ sin 74

b = (3.46 sin 58) / sin 74

b = 3.05 mi.

GL ☺ ☺

MATH QUESTION TRIG!?

The problem below refers to a vector V with magnitude |V| that forms an angle θ with the positive x-axis. In this case, give the magnitudes of the horizontal and vertical vector components of V, namely Vx and Vy, respectively. (Round each answer to the nearest whole number.)
|V| = 371, θ = 12° 40'
|Vx| =
|Vy| =