Respuesta :
       ΔRST ≡ ΔABC
          <R = <A
    (5x + 30)° = (x² - 8x)°
      5x + 30 = x² - 8x
 -x² + 5x + 30 = x² - x² - 8x
 -x² + 5x + 30 = -8x
   + 8x      + 8x
-x² + 13x + 30 = 0
x = -(13) ± √((13)² - 4(-1)(30))
            2(-1)
x = -13 ± √(169 + 120)
           -2
x = -13 ± √(289)
       -2
x = -13 ± 17
      -2
x = -13 + 17Â Â UÂ Â x = -13 - 17
      -2              -2
x = 4Â Â UÂ Â x = -30
    -2         -2
x = -2Â Â Â Â Â x = 25
<C = 4x - 5Â Â Â Â UÂ Â <C = 4x - 5
<C = 4(-2) - 5Â Â UÂ Â <C = 4(25) - 5
<C = -8 - 5Â Â Â Â Â UÂ Â <C = 100 - 5
<C = -13°      U  <C = 95°
or
      ΔRST ≡ ΔABC
    <A + <C = <R
   (x² - 8x)°+ (4x - 5)° = (5x + 30)°
     (x² - 8x) + (4x - 5) = (5x + 30)
     (x² - 8x + 4x - 5) = (5x + 30)
           x² - 4x - 5 = 5x + 30
            - 5x    - 5x    Â
          x² - 9x - 5 = 30
               - 30 - 30
          x² - 9x - 35 = 0
x = -(-9) ± √((-9)² - 4(1)(-35))
             2(1)
x = 9 ± √(81 + 140)
          2
x = 9 ± √(221)
       2
x = 9 ± 14.86
        2
x = 9 + 14.86 Â UÂ Â x = 9 - 14.86
        2              2
x = 23.86Â Â UÂ Â x = -4.14
     2            2
x = 11.93Â Â UÂ Â x = -2.07
          <R = <A
    (5x + 30)° = (x² - 8x)°
      5x + 30 = x² - 8x
 -x² + 5x + 30 = x² - x² - 8x
 -x² + 5x + 30 = -8x
   + 8x      + 8x
-x² + 13x + 30 = 0
x = -(13) ± √((13)² - 4(-1)(30))
            2(-1)
x = -13 ± √(169 + 120)
           -2
x = -13 ± √(289)
       -2
x = -13 ± 17
      -2
x = -13 + 17Â Â UÂ Â x = -13 - 17
      -2              -2
x = 4Â Â UÂ Â x = -30
    -2         -2
x = -2Â Â Â Â Â x = 25
<C = 4x - 5Â Â Â Â UÂ Â <C = 4x - 5
<C = 4(-2) - 5Â Â UÂ Â <C = 4(25) - 5
<C = -8 - 5Â Â Â Â Â UÂ Â <C = 100 - 5
<C = -13°      U  <C = 95°
or
      ΔRST ≡ ΔABC
    <A + <C = <R
   (x² - 8x)°+ (4x - 5)° = (5x + 30)°
     (x² - 8x) + (4x - 5) = (5x + 30)
     (x² - 8x + 4x - 5) = (5x + 30)
           x² - 4x - 5 = 5x + 30
            - 5x    - 5x    Â
          x² - 9x - 5 = 30
               - 30 - 30
          x² - 9x - 35 = 0
x = -(-9) ± √((-9)² - 4(1)(-35))
             2(1)
x = 9 ± √(81 + 140)
          2
x = 9 ± √(221)
       2
x = 9 ± 14.86
        2
x = 9 + 14.86 Â UÂ Â x = 9 - 14.86
        2              2
x = 23.86Â Â UÂ Â x = -4.14
     2            2
x = 11.93Â Â UÂ Â x = -2.07
In any statement like this one:Â [tex]\triangle RST \cong \triangle ABC[/tex] you can assume that the points match up in the order that you are given them.
This means that [tex]\angle A \cong \angle R[/tex].
We know that [tex]m\angle A = x^2-8x[/tex] and [tex]m\angle R=5x+30[/tex], and because they are congruent we can set the two equal to each other.
[tex]x^2-8x=5x+30[/tex]
Let's get everything to one side.
[tex]x^2-13x-30=0[/tex]
Let's solve by factoring, since it's easy to do with these whole numbers.
We're looking for two number thats add to -30 and multiply to -13...
These would be -15 and 2.
Since our leading coefficient (_x²) is 1, we can factor straight to (x-15)(x+2).
Here's what it would look like if you went through all the steps anyways, though.
[tex]x^2-15x+2x-30=0[/tex]
[tex]x(x-15)+2(x-15)=0[/tex]
[tex](x+2)(x-15)=0[/tex]
Any value which causes either factor to equal 0 is a solution.
(The second factor wouldn't matter b/c 0 times anything is still 0)
Therefore x = -2 or 15.
Only one of these is possible, however!
If you use x = -2, you will find that the angle measure 4x-5 is negative, which is impossible. In this case, x must be 15.
Let's find the measure of angle C.
[tex]m\angle C=4x-5\ where\ x=15\\m\angle C=4(15)-5\\m\angle C=60-5\\\boxed{m\angle C = 55\°}[/tex]
This means that [tex]\angle A \cong \angle R[/tex].
We know that [tex]m\angle A = x^2-8x[/tex] and [tex]m\angle R=5x+30[/tex], and because they are congruent we can set the two equal to each other.
[tex]x^2-8x=5x+30[/tex]
Let's get everything to one side.
[tex]x^2-13x-30=0[/tex]
Let's solve by factoring, since it's easy to do with these whole numbers.
We're looking for two number thats add to -30 and multiply to -13...
These would be -15 and 2.
Since our leading coefficient (_x²) is 1, we can factor straight to (x-15)(x+2).
Here's what it would look like if you went through all the steps anyways, though.
[tex]x^2-15x+2x-30=0[/tex]
[tex]x(x-15)+2(x-15)=0[/tex]
[tex](x+2)(x-15)=0[/tex]
Any value which causes either factor to equal 0 is a solution.
(The second factor wouldn't matter b/c 0 times anything is still 0)
Therefore x = -2 or 15.
Only one of these is possible, however!
If you use x = -2, you will find that the angle measure 4x-5 is negative, which is impossible. In this case, x must be 15.
Let's find the measure of angle C.
[tex]m\angle C=4x-5\ where\ x=15\\m\angle C=4(15)-5\\m\angle C=60-5\\\boxed{m\angle C = 55\°}[/tex]