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Trigonometry

Question 7 of 9

Sine and Cosine Rules

Question

Solve triangle \(ABC\) where \(a = 10\), \(b = 12\), \(c = 14\)

Triangle ABC, drawn to scale from the given values. All three sides are given: side c = 14 forms the base from A on the left to B on the right, side b = 12 rises from A to the apex C, and side a = 10 runs from C down to B. No angles are given; all three are to be found.
Show solution Hide solution Fully worked — 7 steps

This is an SSS Case.

  1. Use the cosine rule to solve for \(\angle A\)
    \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \]
  2. Substitute and solve for \(\angle A\)
    \[ \cos A = \frac{12^2 + 14^2 - 10^2}{2(12)(14)} \]
    \[ \cos A = 0.7143 \]
  3. Round the answer to the nearest whole number
    \[ \angle A = 44.41 \approx 44^\circ \]
  4. Use the cosine rule to solve for \(\angle B\)
    \[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]
  5. Substitute and solve for \(\angle B\)
    \[ \cos B = \frac{10^2 + 14^2 - 12^2}{2(10)(14)} \]
    \[ \cos B = 0.5429 \]
    \[ \angle B = 57.12 \approx 57^\circ \]
  6. Solve for \(\angle C\)
    \[ \angle C = 180 - (44 + 57) = 79^\circ \]
  7. Final answer

    Therefore, \(\angle A = 44^\circ\), \(\angle B = 57^\circ\) and \(\angle C = 79^\circ\).

    That is, angle A is 44 degrees, angle B is 57 degrees and angle C is 79 degrees.