Hi, I need help with these problems, I'm not sure where to start. The book's examples are different. Thanks in advance!!
1.) To find the distance AB across a river, a distance BC of 472m is laid off on one side of the river. It is found that B=108.1 degrees and C= 17.9 degrees. Find AB.
2.) Find the angle between the pair of vectors to the nearest tenth of a degree.
9i+2j, -3i+3j
3.) Vector v has the given magnitude an direction. Find the magnitude of the indicated component of v. Find the vertical component of v if:
theta=38.3 degrees. |v|= 281
1. angle A = 180-17.9-108.1 = 54°
now by the sine law:
AB/sin17.9 = 472/sin54
AB = 472sin17.9/sin 54
= ....
2. (9,2)dot((-3,3) = √85√18cosØ , where Ø is the angle between them
(-27 + 6)/(√85√18) = cosØ
I got Ø = 122.5°
sin 38.3 = y/281
y = ....
thank you
Sure! Let's start with problem 1:
To find the distance AB across the river, we can use the law of sines. The law of sines states that in any triangle, the ratio of the length of a side to the sine of the opposite angle is constant.
In this case, we have a triangle ABC, where angle B is 108.1 degrees and angle C is 17.9 degrees. The side opposite angle B is BC, which has a length of 472m.
Let's call the length of side AB as x. According to the law of sines, we can set up the following ratio:
sin(B) / BC = sin(C) / AB
Plugging in the known values:
sin(108.1 degrees) / 472m = sin(17.9 degrees) / x
Now we can solve for x. First, we find the value of sin(108.1 degrees) using a calculator, which is approximately 0.924. Then we have:
0.924 / 472m = sin(17.9 degrees) / x
To solve for x, we can cross-multiply and divide:
0.924 * x = 472m * sin(17.9 degrees)
Simplifying:
x = (472m * sin(17.9 degrees)) / 0.924
Calculate the right side of the equation using a calculator and you will find the value of x, which is the distance AB across the river.
Moving on to problem 2:
To find the angle between two vectors, we can use the dot product. The dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them.
Let's consider the given vectors: vector u = 9i + 2j and vector v = -3i + 3j.
The formula for the dot product is:
u · v = (magnitude of u) * (magnitude of v) * cos(theta)
To find the angle between the two vectors, we need to solve for theta using the dot product formula:
9 * (-3) + 2 * 3 = sqrt((9^2 + 2^2) * ((-3)^2 + 3^2)) * cos(theta)
Simplifying:
-27 + 6 = sqrt(85) * sqrt(18) * cos(theta)
-21 = sqrt(1530) * cos(theta)
To solve for cos(theta), divide both sides by sqrt(1530):
cos(theta) = -21 / sqrt(1530)
Using a calculator, calculate the value of cos(theta) and then find the corresponding angle. Remember to round to the nearest tenth of a degree.
Finally, let's move on to problem 3:
To find the vertical component of vector v, we can use trigonometry. The vertical component is equal to the magnitude of the vector multiplied by the sine of the angle.
In this case, we have a vector v with magnitude |v| = 281 and an angle theta = 38.3 degrees.
The formula for the vertical component is:
Vertical component = |v| * sin(theta)
Plugging in the values:
Vertical component = 281 * sin(38.3 degrees)
Calculate the right side of the equation using a calculator and you will find the magnitude of the indicated vertical component of vector v.
I hope this helps you with your problems! If you have any further questions or need clarification, feel free to ask.