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2 years ago

Applying the Laws of Sines and Cosines

Mathematics

1 answer:

Naddika [18.5K]2 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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Find m< AEB. Please help

MrRissso [65]

Answer:

118

Step-by-step explanation:

the length ED is equal to EC

the angle is also the same

that's mean the angle ECD is equal to 31

then to find the another angle just do like this

=180–31–31

=118

5 0

3 years ago

Read 2 more answers

Find the Value of x so that the ratios are equivalent.

guajiro [1.7K]

x=45

9 x

12 60

x=9×60 divided by 12

9×60=540

540divided by 12=45

7 0

2 years ago

Will mark brainlist

marissa [1.9K]

The answer is 4.4102. We know the length of one side is 2.4 and the length of the other side is half of 7.4. If we use Pythagorean’s theorem to find the hypotenuse we get (7.4/2)^2 + 2.4^2 = X^2. By solving for x we get x= root( (7.4/2)^2 + 2.4^2). So x is approximately 4.41.

8 0

3 years ago

I'll mark brainliest <br>need help ASAP <br>please and thank you

liq [111]

Hello,

For the first question, congruent triangles have congruent corresponding angles and sides. For option choice A, MN corresponds with PQ, not PR, so it is incorrect to conclude that they are equal. The answer is A.

For the second question, we are given two angles and an not included side, so it would be AAS.

The answer is A, I believe.

Hope this helps!

7 0

3 years ago

Y is the midpoint of XZ .X has coordinates (2, 4), and Y has coordinates (-1, 1). Find the coordinates of z.

cricket20 [7]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
X&({{ 2}}\quad ,&{{ 4}})\quad 
% (c,d)
Z&({{ x}}\quad ,&{{ y}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{x+2}{2}~,~\cfrac{y+4}{2} \right)=\stackrel{Y}{(-1,1)}\implies 
\begin{cases}
\cfrac{x+2}{2}=-1\\\\
x+2=-2\\
\boxed{x=-4}\\
------\\
\cfrac{y+4}{2}=1\\\\
y+4=2\\
\boxed{y=-2}
\end{cases}$

7 0

3 years ago