Question

HELp meh with this question it very hard

Answer 1

Answer:

• AB = 7 cm

Step-by-step explanation:

Use the law of cosines to find the side AB:

• $AB = \sqrt{(x + 3)^2+x^2-2x(x+3)cos (60)} =$
• $\sqrt{x^2+6x+9+x^2-x^2-3x} = \sqrt{x^2+3x+9}$

Use the Heron's area formula next:

• $A = \sqrt{s(s - a)(s-b)(s-c)}$, where s- semi perimeter

• s = 1/2[x + x + 3 + $\sqrt{x^2+3x+9}$) = 1/2 (2x + 3 + $\sqrt{x^2+3x+9}$)
• s - a = 1/2 (2x + 3 + $\sqrt{x^2+3x+9}$ - 2x - 6) = 1/2 ($\sqrt{x^2+3x+9 }$ - 3)
• s - b = 1/2 (2x + 3 + $\sqrt{x^2+3x+9}$ - 2x) = 1/2 ($\sqrt{x^2+3x+9}$ + 3)
• s - c = 1/2 (2x + 3 + $\sqrt{x^2+3x+9}$ - 2$\sqrt{x^2+3x+9}$) = 1/2 (2x + 3 - $\sqrt{x^2+3x+9}$)

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:

• $AB = \sqrt{40 + 9} = 7 cm$