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3 years ago

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For ΔABC, ∠A = 3x, ∠B = 2x - 3, and ∠C = x + 3. If ΔABC undergoes a dilation by a scale factor of 2 to create ΔA'B'C' with ∠A' =

2x + 30, ∠B' = x + 27, and ∠C' = 1/2x + 18, which confirms that ΔABC∼ΔA'B'C by the AA criterion?
A) ∠A = ∠A' = 75° and ∠B = ∠B' = 47°
B) ∠A = ∠A' = 60° and ∠B = ∠B' = 37°
C) ∠B = ∠B' = 57° and ∠C = ∠C' = 33°
D) ∠A = ∠A' = 105° and ∠C = ∠C' = 38°

2 answers:

Mashutka [201]3 years ago

3 0

I am not 100% sure but I think the answer is C I hope I helped you and Good luck

ioda3 years ago

3 0

Answer: C) ∠B = ∠B' = 57° and ∠C = ∠C' = 33°

Step-by-step explanation:

GIven:- In ΔABC, ∠A = 3x, ∠B = 2x - 3, and ∠C = x + 3.

In ΔA'B'C' ∠A' = 2x + 30, ∠B' = x + 27, and ∠C' = 1/2x + 18

By angle sum property in ΔABC

$\Rightarrow3x+2x-3+x+3=180^{\circ}\\\Rightarrow6x=180^{\circ}\\\Rightarrow\ x=30^{\circ}$

⇒ ∠A =3(30)=90°

∠B=2(30)-3=57°

∠C=30+3=33°

By angle sum property in ΔA'B'C'

$\Rightarrow2x+x+27+\frac{1}{2}x+18=180^{\circ}\\\Rightarrow\frac{7}{2}x+75=180^{\circ}\\\Rightarrow7x+150=180^{\circ}\\\Rightarrow7x=210^{\circ}\\\Rightarrow\ x=30^{\circ}$

∠A'=2(30)+30=90°

∠B'=30+27=57°

∠C'= $\frac{1}{2}(30)+18=33^{\circ}$

therefore, C) ∠B = ∠B' = 57° and ∠C = ∠C' = 33° gives ΔABC∼ΔA'B'C by the AA criterion.

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$\displaystyle x= -\frac{10}{\sqrt{269}}\\\\\displaystyle y= \frac{13}{\sqrt{269}}\\\\\displaystyle z = \frac{23\sqrt{269}+269}{269}$

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Step-by-step explanation:

<u>Lagrange Multipliers</u>

It's a method to optimize (maximize or minimize) functions of more than one variable subject to equality restrictions.

Given a function of three variables f(x,y,z) and a restriction in the form of an equality g(x,y,z)=0, then we are interested in finding the values of x,y,z where both gradients are parallel, i.e.

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for some scalar $\lambda$ called the Lagrange multiplier.

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$\bigtriangledown f=\lambda \bigtriangledown g+\mu \bigtriangledown h$

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$\bigtriangledown f=$

Considering each variable as independent we have three equations right from the Lagrange condition, plus one for each restriction, to form a 5x5 system of equations in $x,y,z,\lambda,\mu$ .

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$f(x, y, z) = x + 2y + 11z\\g(x, y, z) = x - y + z -1=0\\h(x, y, z) = x^2 + y^2 -1= 0$

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$f_x=1\ ,f_y=2\ ,f_z=11\ \\g_x=1\ ,g_y=-1\ ,g_z=1\\h_x=2x\ ,h_y=2y\ ,h_z=0$

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$1=\lambda (1)+\mu (2x)\\2=\lambda (-1)+\mu (2y)\\11=\lambda (1)+\mu (0)$

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And for

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Replacing into f, we get

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Answer:

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Step-by-step explanation:

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