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sergey [27]
4 years ago
12

Triangle ABC has been translated to create triangle A'B'C'. Angles C and C' are both 32 degrees, angles B and B' are both 72 deg

rees, and sides BC and B'C' are both 5 units long. Which postulate or theorem below would prove the two triangles are congruent?

Mathematics
2 answers:
Murrr4er [49]4 years ago
8 0
Δ ABC and Δ A'B'C' are congruent.

Proof: BC = B'C' = 5 units (given)

Angle B = Angle B' = 72° (given)
Angle C = Angle C' = 32° (given)

The 2 tringles are ten equal ASA

Karo-lina-s [1.5K]4 years ago
6 0

Answer:

ΔABC ≅ ΔA'B'C' are congruent by ASA postulate

Step-by-step explanation:

Let's draw ΔABC and ΔA'B'C'

<u>Statement                                                 Reason</u>

i) ∠C = ∠C' = 32°                                            Given

ii) ∠B = ∠B' = 72°                                            Given

iii) BC = B'C' = 5                                              Given

IV) ΔABC ≅ ΔA'B'C'                                        If two angles and the included

                                                                        side of one triangle equal to the

                                                                        corresponding angles and

                                                                        included side of another triangle

                                                                        then the two triangles are

                                                                        congruent. By ASA postulate.

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X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square
swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = -\frac{4}{5}x+\frac{58}{5}

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

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   (x + 2)² + (y - 5)²- 9 = 30

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By comparing this equation with the standard equation of a circle,

    Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

    Slope of the line joining this point to the center (-2, 5),

    m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

          = \frac{10-5}{2+2}

          = \frac{5}{4}

    Let the slope of the tangent which is perpendicular to this line is 'm_{2}'

    Then by the property of perpendicular lines,

          m_{1}\times m_{2}=-1

          \frac{5}{4}\times m_{2}=-1

                 m_{2}=-\frac{4}{5}

   Now the equation of the line passing though (2, 10) having slope m_{2}=-\frac{4}{5}

           y - y' = m_{2}(x-x')

           y - 10 = -\frac{4}{5}(x-2)

           y - 10 = -\frac{4}{5}x+\frac{8}{5}

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Therefore, equation of the line will be, y = -\frac{4}{5}x+\frac{58}{5}

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