Standard High contrast

Trigonometry

Question 2 of 9

Area of Triangle

Question

Find the area of \(\triangle ABC\) whose \(\angle A = 35^\circ\), \(\angle B = 57^\circ\), \(a = 11.45\,\text{cm}\) and \(b = 17\,\text{cm}\) (Round your answer to 2 decimal places)

Show solution Hide solution Fully worked — 4 steps
Triangle ABC: vertex A at the bottom left, vertex B at the top right and vertex C at the bottom right, almost directly below B. The angle at A is marked 35 degrees and the angle at B is marked 57 degrees. Side b, from A to C along the base, is 17 cm; side a, from B down to C, is 11.45 cm; side c, from A to B, is unlabelled. The solution finds the third angle C, then uses the two given sides a and b and their included angle C.
  1. State the formula
    \[ \mathit{area} = \frac{1}{2} ab \sin C \]
  2. Find the value of angle C
    \[ C = 180^\circ - (35^\circ + 57^\circ) \]

    Add the 2 given angles:

    \[ C = 180^\circ - 92^\circ \]

    Subtract the sum of the 2 angles from 180:

    \[ C = 88^\circ \]
  3. Substitute values based on formula
    \[ \mathit{area} = \frac{1}{2}(11.45\,\text{cm})(17\,\text{cm})(\sin 88^\circ) \]
  4. Final answer

    Evaluate \(\frac{1}{2}(11.45\,\text{cm})(17\,\text{cm})(\sin 88^\circ)\):

    \[ \mathit{area} = 97.27\,\text{cm}^2 \]

    That is, the area of triangle ABC is 97.27 square centimetres, to 2 decimal places.