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Trigonometry

Question 3 of 9

Sine and Cosine Rules

Question

Solve triangle \(ABC\) given \(a = 28\), \(b = 15\), and \(\angle A = 110^\circ\). Find \(c\), \(\angle B\) and \(\angle C\). (Round your answer to the nearest whole number)

Show solution Hide solution Fully worked — 7 steps

Condition for one solution: \(\angle A\) is obtuse and \(a \gt b\).

  1. Use the sine rule to solve for \(\angle B\)
    \[ \frac{a}{\sin A} = \frac{b}{\sin B} \]

    Substitute the values of \(a\), \(b\) and \(\angle A\):

    \[ \frac{28}{\sin 110^\circ} = \frac{15}{\sin B} \]
  2. Cross multiply and solve for \(\sin B\)
    \[ \sin B = \frac{15 \sin 110}{28} = 0.5034 \]
  3. Solve for angle \(B\)
    \[ \angle B = 30.23^\circ \approx 30^\circ \]
  4. Solve for angle \(C\)

    The sum of the angles of a triangle is \(180^\circ\):

    \[ \angle C = 180 - (30.23 + 110) = 39.77 \approx 40^\circ \]
  5. Use the sine rule to solve for \(c\)
    \[ \frac{a}{\sin A} = \frac{c}{\sin C} \]

    Substitute the values of \(a\), \(\angle A\) and \(\angle C\):

    \[ \frac{28}{\sin 110^\circ} = \frac{c}{\sin 39.77^\circ} \]
  6. Cross multiply and solve for \(c\)
    \[ c = \frac{28 \sin 39.77}{\sin 110} \]
    \[ c = 19.06 \approx 19 \]
  7. Final answer

    Therefore, \(c = 19\), \(\angle B = 30^\circ\) and \(\angle C = 40^\circ\). That is, side c is 19, angle B is 30 degrees, and angle C is 40 degrees.