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Trigonometry

Question 6 of 9

Sine and Cosine Rules

Question

Solve the given triangle where \(a = 25\), \(b = 40\) and \(\angle C = 60^\circ\)

Triangle ABC, drawn to scale from the given values. Side b = 40 runs from A up to C, side a = 25 runs from C down to B, and the included angle between them at C, marked by the arc, is 60°. The unmarked arcs at A and B are the angles to be found; the third side, AB (side c), is unlabelled and is also to be found.
Show solution Hide solution Fully worked — 7 steps

This is an SAS Case.

  1. Use the cosine rule to solve for \(c\)
    \[ c^2 = a^2 + b^2 - 2ab\cos C \]
  2. Substitute values to solve for \(c\)
    \[ c^2 = 25^2 + 40^2 - 2(25)(40)\cos 60^\circ \]
    \[ c^2 = 1225 \]
    \[ c = 35 \]
  3. Use the cosine rule to solve for \(\angle A\)
    \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
  4. Substitute values to solve for angle A
    \[ \cos A = \frac{40^2 + 35^2 - 25^2}{2(40)(35)} \]
    \[ \cos A = 0.7857 \]
  5. Round the answer to the nearest whole number
    \[ \angle A = 38.21 \approx 38^\circ \]
  6. Solve for angle B
    \[ \angle B = 180 - (38 + 60) = 82^\circ \]
  7. Final answer

    Therefore, \(c = 35\), \(\angle A = 38^\circ\) and \(\angle B = 82^\circ\).

    That is, side c is 35, angle A is 38 degrees and angle B is 82 degrees.