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Sagot :
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:
https://brainly.com/question/27088353
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