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Trigonometry

Question 8 of 9

Area of Triangle

Question

What would be the area of a triangle having angles \(40^\circ\), \(30^\circ\) and \(110^\circ\) respectively, and a \(39.2\,\text{cm}\) side opposite to the biggest angle? (Round your answer to 2 decimal places)

Show solution Hide solution Fully worked — 4 steps
Triangle ABC, drawn to scale, with vertices A at the top left, B at the top right and C at the bottom. All three angles are marked with arcs and their values: 40 degrees at A, 30 degrees at B and 110 degrees at C. The given side c = 39.2, opposite the biggest angle at C, runs along the top from A to B. Sides a and b are unlabelled.
  1. State the formula
    \[ \mathit{area} = \frac{1}{2}\,bc \sin A \]
  2. Use sine rule to get the value of \(b\)
    \[ \frac{\sin 30^\circ}{b} = \frac{\sin 110^\circ}{39.2} \]

    Cross multiply to solve for \(b\):

    \[ b = \frac{(39.2)(\sin 30^\circ)}{\sin 110^\circ} \]

    Evaluate \(\dfrac{(39.2)(\sin 30^\circ)}{\sin 110^\circ}\):

    \[ b = 20.86\,\text{cm} \]
  3. Substitute values based on formula
    \[ \mathit{area} = \frac{1}{2}(20.86)(39.2)(\sin 40^\circ) \]
  4. Final answer

    Evaluate \(\dfrac{1}{2}(20.86)(39.2)(\sin 40^\circ)\):

    \[ \mathit{area} = 262.81\,\text{cm}^2 \]

    That is, the area of the triangle is 262.81 square centimetres, to 2 decimal places.