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Difficulty: HardSine and Cosine Rules

In ΔABC\Delta ABC, the lengths of the sides are a=x cma = x\text{ cm}, b=(x+3) cmb = (x + 3)\text{ cm}, and c=(x+2) cmc = (x + 2)\text{ cm}. If ∠C=60∘\angle C = 60^\circ, what is the value of xx?

  1. A
    3
  2. 5Answer
  3. C
    7
  4. D
    1

Answer

5
Applying the Cosine Rule c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C with a=xa = x, b=x+3b = x + 3, c=x+2c = x + 2, and cos⁡60∘=12\cos 60^\circ = \frac{1}{2} gives (x+2)2=x2+(x+3)2−x(x+3)(x+2)^2 = x^2 + (x+3)^2 - x(x+3). Expanding both sides leads to x2+4x+4=x2+3x+9x^2 + 4x + 4 = x^2 + 3x + 9. Subtracting x2x^2 and isolating xx yields x=5x = 5.

Step-by-Step Solution

1
Apply the Cosine Rule for side cc and angle CC.
c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C
The Cosine Rule relates all three sides of a triangle to the cosine of an included angle.
2
Substitute a=xa = x, b=x+3b = x + 3, c=x+2c = x + 2, and cos⁡60∘=12\cos 60^\circ = \frac{1}{2} into the formula.
(x+2)2=x2+(x+3)2−2(x)(x+3)(12)(x + 2)^2 = x^2 + (x + 3)^2 - 2(x)(x + 3)\left(\frac{1}{2}\right)
Inserting the given algebraic side lengths and angle allows solving for xx.
3
Expand both sides of the equation.
x2+4x+4=x2+(x2+6x+9)−(x2+3x)x^2 + 4x + 4 = x^2 + (x^2 + 6x + 9) - (x^2 + 3x)
Cancel the factor of 22 with 12\frac{1}{2} and expand (x+2)2(x+2)^2 and (x+3)2(x+3)^2.
4
Simplify the right-hand side.
x2+4x+4=x2+3x+9x^2 + 4x + 4 = x^2 + 3x + 9
Combine like terms: (x2+x2−x2)+(6x−3x)+9=x2+3x+9(x^2 + x^2 - x^2) + (6x - 3x) + 9 = x^2 + 3x + 9.
5
Subtract x2x^2 from both sides and isolate xx.
4x−3x=9−4  ⟹  x=54x - 3x = 9 - 4 \implies x = 5
Subtracting x2+3x+4x^2 + 3x + 4 from both sides gives the linear solution x=5x = 5.

Key Concept

Solving for unknown algebraic side lengths using the Cosine Rule.
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Topics

Sine and Cosine Rules

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