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Trigonometry

Question 3 of 9

Area of Triangle

Question

Find the area of \(\triangle ABC\) whose \(\angle A = 20^\circ\), \(\angle B = 55^\circ\) and \(c = 14.14\,\text{cm}\) (Round your answer to 2 decimal places)

Show solution Hide solution Fully worked — 5 steps
Triangle ABC: vertex A at the bottom left, vertex C at the bottom right and vertex B above and slightly beyond C. The angle at A is marked 20 degrees and the angle at B is marked 55 degrees. Side c, from A up to B, is 14.14 cm; side b (from A to C along the base) and side a (from B down to C) are unlabelled. The solution finds the third angle C, uses the sine rule to find side a, then uses sides a and c and their included angle B.
  1. State the formula
    \[ \mathit{area} = \frac{1}{2} ac \sin B \]
  2. Subtract the sum of the 2 angles from 180 to get C
    \[ C = 180^\circ - (20^\circ + 55^\circ) \]
    \[ C = 105^\circ \]
  3. Use sine rule to get side a
    \[ \frac{\sin 20^\circ}{a} = \frac{\sin 105^\circ}{14.14} \]

    Cross multiply to solve for \(a\):

    \[ a = \frac{14.14(\sin 20^\circ)}{\sin 105^\circ} \]

    Evaluate \(\frac{14.14(\sin 20^\circ)}{\sin 105^\circ}\):

    \[ a = 5.01\,\text{cm} \]
  4. Substitute values based on formula
    \[ \mathit{area} = \frac{1}{2}(5.01\,\text{cm})(14.14\,\text{cm})(\sin 55^\circ) \]
  5. Final answer

    Evaluate \(\frac{1}{2}(5.01\,\text{cm})(14.14\,\text{cm})(\sin 55^\circ)\):

    \[ \mathit{area} = 29.01\,\text{cm}^2 \]

    That is, the area of triangle ABC is 29.01 square centimetres, to 2 decimal places.