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vovikov84 [41]
2 years ago
8

Triangle ABC has vertices A (1,1), B (3,2) and C (0,-2), It is dilated by a scale factor of 1/2 about the origin to create trian

gle A' B' C'. What is the length in units of side B'C'?
Mathematics
1 answer:
Ghella [55]2 years ago
8 0

Answer:

The length is 2.5 units

Step-by-step explanation:

Before we find the value of the length of the side, we need the measure of the coordinates after dilation

B’ = 1/2 B = 1/2(3,2) = (1.5, 1)

C’ = 1/2 C = 1/2( 0,-2) = (0,-1)

The measure of the length is the distance between the two points

Mathematically, that would be;

D = √(x2-x1)^2 + (y2-y1)^2

D = √(0-1.5)^2 + (-1-1)^2

D = √(2.25 + 4)

D = √6.25

D = 2.5 units

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I take it we aren't solving for x.


So


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2 years ago
drawbridge at the entrance to an ancient castle is raised and lowered by a pair of chains. The figure represents the drawbridge
jasenka [17]

Answer:

4.0 meters, ∠C = 39°, ∠A = 51°

Step-by-step explanation:

Firstly, our diagram shows us that the given triangle is actually a right triangle. So we can use the <em>Pythagorean Theorem</em> to solve for the height of the chain:

a^{2} +b^{2} =c^{2}

(3.3)^{2} +b^{2} =(5.2)^{2}

b^{2} =(5.2)^{2}-(3.3)^{2}

b =\sqrt{(5.2)^{2}-(3.3)^{2}}

b=4.0187...

b=4.0 m

Now, we can use the <em>Law of Cosines</em> to figure out one of the acute angles:

c^{2}  =a^{2} +b^{2} -2ab(cos(C))

(3.3)^{2}  =(4.0)^{2} +(5.2)^{2} -2(4.0)(5.2)(cos(C))

cos(C)=\frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)}

C=cos^{-1}( \frac{(3.3)^{2}-(4.0)^{2} -(5.2)^{2}}{-2(4.0)(5.2)})

C=39.3915...

∠C = 39°

And since we know that all angles in a triangle add up to 180°:

∠A + ∠B + ∠C = 180

∠A + 90 + 39 = 180

∠A = 180 - 90 - 39

∠A = 51°

However, you should always review any answers on the Internet and make sure they are correct! Check my work to see if I made any mistakes!

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Step-by-step explanation:

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-2 and 1.

Hope this helps!
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