Applying the Laws of Sines and Cosines

Mathematics

1 answer:

Naddika [18.5K]3 years ago

4 0

Essentially, you would take the formula for the laws of sines and cosines: a^2=b^2+c^2-2bc(Cos(theta)). a is the unknown side you are solving for and theta is the opposite angle to a. Since the angle opposite to R is supplementary to the 50 degree angle, 180-50=130. b could be assigned to either of the given sides; the same goes for c (e.g. b=40 and c=38). Thus, the question should be straightforward in plugging in quantities and solving for the resultant force.

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What’s equivalent (2x^5)^3

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Answer:

8x^15

Step-by-step explanation:

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3 years ago

Find the work done by the force Bold Upper F equals xy Bold i plus (y minus x )Bold j over the straight line from (negative 1 co

aalyn [17]

Answer:

<h2>4/3 Joules </h2>

Step-by-step explanation:

Work is said to be done when force applied to an object causes the object to move through a distance.

Work done = Force * perpendicular distance.

$\int\limits^a_b {F} \, ds$

Given Force F = xy i + (y-x) j and a straight line (-1, -2) to (1, 2)

First we need to get the equation of the straight line given.

Given the slope intercept form y = mx+c

m is the slope

c is the intercept

m = y₂-y₁/x₂-x₁

m = 2-(-2)/1-(-1)

m = 4/2

m = 2

To get the slope we will substtutte any f the point and the slope into the formula y = mx+c

Using the point (1,2)

2 = 2+c

c = 0

y = 2x

Substituting y = 2x into the value of the force F = xy i + (y-x) j we will have;

F = x(2x) i + (2x - x) j

Using the coordinate (1, 2) as the value of s

$W = \int\limits^a_b ({2x^2 i + x j}) \, (i+2j)\\W = \int\limits^a_b ({2x^{2}+2x }) \, dx \\W = [\frac{2x^{3} }{3} +x^{2} ]\left \ x_2=1} \atop {x_1=-1}} \right.\\W = (2(1)^3/3 + 1^2) - (2(-1)^3/3 + (-1)^2)\\W =(2/3+1) - (-2/3+1)\\W = 2/3+2/3+1-1\\W = 4/3 Joules$