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Answer :

Answer:

  • AB = 7 cm

Step-by-step explanation:

Use the law of cosines to find the side AB:

  • [tex]AB = \sqrt{(x + 3)^2+x^2-2x(x+3)cos (60)} =[/tex]
  • [tex]\sqrt{x^2+6x+9+x^2-x^2-3x} = \sqrt{x^2+3x+9}[/tex]

Use the Heron's area formula next:

  • [tex]A = \sqrt{s(s - a)(s-b)(s-c)}[/tex], where s- semi perimeter
  • s = 1/2[x + x + 3 + [tex]\sqrt{x^2+3x+9}[/tex]) = 1/2 (2x + 3 + [tex]\sqrt{x^2+3x+9}[/tex])
  • s - a = 1/2 (2x + 3 + [tex]\sqrt{x^2+3x+9}[/tex] - 2x - 6) = 1/2 ([tex]\sqrt{x^2+3x+9 }[/tex] - 3)
  • s - b = 1/2 (2x + 3 + [tex]\sqrt{x^2+3x+9}[/tex] - 2x) = 1/2 ([tex]\sqrt{x^2+3x+9}[/tex] + 3)
  • s - c = 1/2 (2x + 3 + [tex]\sqrt{x^2+3x+9}[/tex] - 2[tex]\sqrt{x^2+3x+9}[/tex]) = 1/2 (2x + 3 - [tex]\sqrt{x^2+3x+9}[/tex])

Now

  • (s - a)(s - b) = 1/4 [(x²+3x+9) - 9] = 1/4 (x² + 3x)
  • s(s - c) = 1/4 [(2x + 3)² - (x² + 3x + 9)] = 1/4 (3x²+ 9x) = 3/4(x² + 3x)

Next

  • A² = 3/16(x² + 3x)(x² + 3x)
  • 300 = 3/16(x² + 3x)²
  • 1600 = (x² + 3x)²
  • x² + 3x = 40

Substitute this into the first equation:

  • [tex]AB = \sqrt{40 + 9} = 7 cm[/tex]

Answer:

AB = 7 cm

Step-by-step explanation:

Sine Rule for Area

[tex]\sf Area =\dfrac{1}{2}ab \sin C[/tex]

where:

  • a, b and c are the sides opposite angles A, B and C
  • a and b are the sides and C is the included angle

Given:

  • Area = √(300) cm²
  • a = (x + 3) cm
  • b = x cm
  • C = 60°

Substitute the given values into the formula and solve for x:

[tex]\begin{aligned} \sf Area & = \dfrac{1}{2}ab \sin C \\\\\implies \sqrt{300} & = \dfrac{1}{2}(x+3)x \sin 60^{\circ}\\\\\sqrt{300} & = \dfrac{\sqrt{3}}{4}(x+3)x\\\\\dfrac{4\sqrt{300}}{\sqrt{3}} & = x^2+3x\\\\40 & = x^2+3x\\\\x^2+3x-40 & = 0\\\\ (x-5)(x+8)& = 0\\\\ \implies x & = 5, -8\end{aligned}[/tex]

As length is positive, x = 5 only.

Substitute the found value of x into the expressions for the side lengths:

  • a = 5 + 3 = 8 cm
  • b = 5 cm

Cosine rule

[tex]c^2=a^2+b^2-2ab \cos C[/tex]

(where a, b and c are the sides and C is the angle opposite side c)

Substitute the found values into the formula and solve for AB:

[tex]\begin{aligned}c^2 & =a^2+b^2-2ab \cos C\\\implies AB^2 & =8^2+5^2-2(8)(5) \cos 60^{\circ}\\AB^2 & =89-40\\AB^2 & =49\\AB & =\sqrt{49}\\AB & = \pm 7\end{aligned}[/tex]

As length is positive, AB = 7 cm.

Learn more about the sine rule for area here:

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