Standard High contrast

Trigonometry

Question 9 of 9

Area of Triangle

Question

In triangle ABC, \(a = (x+2)\), \(c = (2x+2)\) and \(B = 25^\circ\). If the area of the triangle is \(1\,\text{m}^2\), what is the value of \(x\)? (Round your answer to 2 decimal places)

Show solution Hide solution Fully worked — 6 steps
Triangle ABC, drawn to scale from the solved value of x, with vertices A at the top left, B at the top right and C at the bottom. The angle at B is marked with an arc labelled 25 degrees. The side from A to B, along the top, is labelled 2x + 2, and the side from B down to C is labelled x + 2. The side from A to C is unlabelled.
  1. State the formula
    \[ \mathit{area} = \frac{1}{2}\,ac \sin B \]
  2. Substitute values based on formula
    \[ 1\,\text{m}^2 = \frac{1}{2}(x + 2)(2x + 2)(\sin 25^\circ) \]
  3. Factor and simplify terms
    \[ 1 = \frac{1}{2}(x + 2)(2)(x + 1)(\sin 25^\circ) \]
  4. Rearrange into a quadratic equation

    Divide both sides by \(\sin 25^\circ\):

    \[ \frac{1}{\sin 25^\circ} = (x + 2)(x + 1) \]

    Multiply \((x + 2)\) and \((x + 1)\):

    \[ \frac{1}{\sin 25^\circ} = x^2 + 3x + 2 \]

    Divide 1 by \(\sin 25^\circ\):

    \[ 2.37 = x^2 + 3x + 2 \]

    Subtract 2.37 from both sides:

    \[ 0 = x^2 + 3x - 0.37 \]
  5. Use quadratic formula to get \(x\)
    \[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-0.37)}}{2(1)} \]

    Evaluate \(3^2 - 4(1)(-0.37)\) and \((2)(1)\), taking the positive root:

    \[ x = \frac{-3 + \sqrt{10.48}}{2} \]

    Take the square root of 10.48:

    \[ x = \frac{-3 + 3.24}{2} \]

    Add \(-3\) and 3.24:

    \[ x = \frac{0.24}{2} \]
  6. Final answer

    Divide 0.24 by 2:

    \[ x = 0.12 \]

    That is, the value of \(x\) is 0.12, to 2 decimal places.